Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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in a class of 69 students 41 are taking french , 22 are taking german . of the students taking french or german , 9 are taking both courses . how many students are not enrolled in either course ? | "formula for calculating two overlapping sets : a + b - both + not ( a or b ) = total so in our task we have equation : 41 ( french ) + 22 ( german ) - 9 ( both ) + not = 69 54 + not = 69 not = 69 - 54 = 15 so answer is b" | a ) 6 , b ) 15 , c ) 24 , d ) 33 , e ) 54 | b | subtract(69, subtract(add(41, 22), 9)) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)| | other |
the charge for a single room at hotel p is 50 percent less than the charge for a single room at hotel r and 20 percent less than the charge for a single room at hotel g . the charge for a single room at hotel r is what percent greater than the charge for a single room at hotel g ? | "p = 0.5 r = 0.8 g r = 0.8 g / 0.5 = 1.6 * g thus r is 60 % greater than g . the answer is b ." | a ) 50 % , b ) 60 % , c ) 75 % , d ) 80 % , e ) 90 % | b | multiply(divide(subtract(const_100, multiply(divide(subtract(const_100, 50), subtract(const_100, 20)), const_100)), multiply(divide(subtract(const_100, 50), subtract(const_100, 20)), const_100)), const_100) | subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)|divide(#4,#3)|multiply(#5,const_100)| | gain |
laxmi and prasanna set on a journey . laxmi moves northwards at a speed of 18 kmph and prasanna moves southward at a speed of 27 kmph . how far will be prasanna from laxmi after 60 minutes ? | "explanation : we know 60 min = 1 hr total northward laxmi ' s distance = 18 kmph x 1 hr = 18 km total southward prasanna ' s distance = 27 kmph x 1 hr = 27 km total distance between prasanna and laxmi is = 18 + 27 = 45 km . answer : d" | a ) 11 , b ) 50 , c ) 28 , d ) 45 , e ) 18 | d | add(18, 27) | add(n0,n1)| | physics |
a sum of money at simple interest amounts to rs . 815 in 3 years and to rs . 894 in 4 years . the sum is : | "s . i . for 1 year = rs . ( 894 - 815 ) = rs . 79 . s . i . for 3 years = rs . ( 79 x 3 ) = rs . 237 . principal = rs . ( 815 - 237 ) = rs . 578 . answer : option e" | a ) s . 650 , b ) s . 690 , c ) s . 698 , d ) s . 700 , e ) s . 578 | e | subtract(815, divide(multiply(subtract(894, 815), 3), 4)) | subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)| | gain |
tom traveled the entire 80 miles trip . if he did the first 30 miles of at a constant rate 30 miles per hour and the remaining trip of at a constant rate 50 miles per hour , what is the his average speed , in miles per hour ? | "avg speed = total distance / total time = ( d 1 + d 2 ) / ( t 1 + t 2 ) = ( 30 + 50 ) / ( ( 30 / 30 ) + ( 50 / 50 ) ) = 80 / 2 = 40 mph c" | a ) 55 mph , b ) 50 mph , c ) 40 mph , d ) 60 mph , e ) 70 mph | c | divide(80, add(divide(50, subtract(80, 30)), divide(30, 30))) | divide(n1,n2)|subtract(n0,n1)|divide(n3,#1)|add(#2,#0)|divide(n0,#3)| | physics |
there are 24 students in a seventh grade class . they decided to plant birches and roses at the school ' s backyard . while each girl planted 3 roses , every three boys planted 1 birch . by the end of the day they planted 2424 plants . how many birches and roses were planted ? | let xx be the number of roses . then the number of birches is 24 β x 24 β x , and the number of boys is 3 Γ ( 24 β x ) 3 Γ ( 24 β x ) . if each girl planted 3 roses , there are x 3 x 3 girls in the class . we know that there are 24 students in the class . therefore x 3 + 3 ( 24 β x ) = 24 x 3 + 3 ( 24 β x ) = 24 x + 9 ... | a ) 6 , b ) 7 , c ) 8 , d ) 2 , e ) 9 | a | divide(subtract(multiply(3, 24), 24), subtract(multiply(3, 3), 1)) | multiply(n0,n1)|multiply(n1,n1)|subtract(#0,n0)|subtract(#1,n2)|divide(#2,#3) | physics |
having received his weekly allowance , a student spent 3 / 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 0.40 at the candy store . what is this student β s weekly allowance ? | "let x be the value of the weekly allowance . ( 2 / 3 ) ( 2 / 5 ) x = 40 cents ( 4 / 15 ) x = 40 x = $ 1.50 the answer is b ." | a ) $ 1.20 , b ) $ 1.50 , c ) $ 1.80 , d ) $ 2.00 , e ) $ 2.50 | b | divide(multiply(multiply(3, 5), 0.40), const_4) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_4)| | general |
3 / 4 of 5 / 7 of a number is greater than 2 / 3 of 4 / 7 of the same number by 26 . what is half of that number ? | "let no . be x 3 / 4 * 5 / 7 * x - 2 / 3 * 4 / 7 * x = 26 by further solving 15 x / 28 - 8 x / 21 = 26 13 x / 84 = 26 x = 168 we have to find x / 2 = 168 / 2 = 84 answer : d" | a ) 32 , b ) 78 , c ) 92 , d ) 84 , e ) 96 | d | divide(divide(26, subtract(multiply(divide(3, 4), divide(5, 7)), multiply(divide(4, 5), divide(3, 3)))), 4) | divide(n0,n1)|divide(n2,n3)|divide(n6,n2)|divide(n0,n5)|multiply(#0,#1)|multiply(#2,#3)|subtract(#4,#5)|divide(n8,#6)|divide(#7,n6)| | general |
robert left from a pvt company . management hold his salary rs . 15000 / - for one month . earlier robert earned a performance incentive rs . 7280 / - from company . but robert forgot that . after one month robert asked his salary and accountant gives rs . 18500 / - to him . what is the bonus amount given to robert ? | total salary = rs . 15000 / - incentive earned earlier = 7280 / - balance salary = 15000 - 7280 = 7720 paid amount = 18500 / - bonus = 18500 - 7720 = 10780 / - answer is c | a ) a ) 9500 , b ) b ) 12500 , c ) c ) 10780 , d ) d ) 10500 , e ) e ) 8600 | c | subtract(18500, 7280) | subtract(n2,n1) | general |
if a and b run on a circular path in opposite directions of constant speeds with different velocities from p and q . first time they meet 800 mts away from p and second time they meet 700 mts away from p , what is the lenght of the circular path . | 800 + 800 + 700 = 2300 mtrs answer : a | a ) 2300 mtrs , b ) 2400 mtrs , c ) 2500 mtrs , d ) 2600 mtrs , e ) 2700 mtrs | a | add(multiply(800, const_2), 700) | multiply(n0,const_2)|add(n1,#0) | physics |
the annual interest rate earned by an investment increased by 10 percent from last year to this year . if the annual interest rate earned by the investment this year was 11 percent , what was the annual interest rate last year ? | "let us suppose the interest rate last year is x , this year the interest rate is increased by 10 % . hence this year interest rate would be 1.1 * x . setting up the equation - - - - - > 11 = 1.1 * x x = 10 % answer : d" | a ) 1 % , b ) 1.1 % , c ) 9.1 % , d ) 10 % , e ) 10.8 % | d | divide(multiply(11, const_100), add(11, const_100)) | add(n1,const_100)|multiply(n1,const_100)|divide(#1,#0)| | gain |
the cost of the paint is rs . 36.50 per kg . if 1 kg of paint covers 16 squares feet , how much will it cost to paint outside of a cube having 8 feet each side ? | "solution surface area of the cube = ( 6 Γ 82 ) sq . ft = 384 sq . ft . quantity of paint required = ( 384 / 16 ) kg βΉ = βΊ 24 kg . cost of painting = rs . ( 36.50 Γ 24 ) = rs . 876 . answer c" | a ) rs . 692 , b ) rs . 768 , c ) rs . 876 , d ) rs . 972 , e ) none | c | multiply(divide(surface_cube(8), 16), 36.50) | surface_cube(n3)|divide(#0,n2)|multiply(n0,#1)| | geometry |
express a speed of 84 kmph in meters per second ? | "84 * 5 / 18 = 23.33 mps answer : b" | a ) 13.33 mps , b ) 23.33 mps , c ) 33.33 mps , d ) 25.33 mps , e ) 43.33 mps | b | multiply(const_0_2778, 84) | multiply(n0,const_0_2778)| | physics |
a shopkeeper sells 300 metres of cloth for rs . 9000 at a loss of rs . 6 per metre . find his cost price for one metre of cloth ? | "sp per metre = 9000 / 300 = rs . 30 loss per metre = rs . 6 cp per metre = 30 + 6 = rs . 36 answer : c" | a ) s . 59 , b ) s . 58 , c ) s . 36 , d ) s . 46 , e ) s . 13 | c | add(divide(9000, 300), 6) | divide(n1,n0)|add(n2,#0)| | gain |
the h . c . f . of two numbers is 42 and the other two factors of their l . c . m . are 12 and 14 . the larger of the two numbers is : | "explanation : clearly , the numbers are ( 42 x 12 ) and ( 42 x 14 ) . { \ color { blue } \ therefore } larger number = ( 42 x 14 ) = 588 . answer : a ) 588" | a ) 588 , b ) 642 , c ) 567 , d ) 344 , e ) 789 | a | multiply(42, 14) | multiply(n0,n2)| | other |
jack , jill , and sandy each have one try to make a basket from half court . if their individual probabilities of making the basket are 1 / 6 , 1 / 7 , and 1 / 8 respectively , what is the probability that all three will make a basket ? | "the probability that all three will make a basket is 1 / 6 * 1 / 7 * 1 / 8 = 1 / 336 . the answer is a ." | a ) 1 / 336 , b ) 5 / 336 , c ) 11 / 336 , d ) 1 / 168 , e ) 1 / 84 | a | multiply(multiply(divide(1, 7), divide(1, 8)), subtract(1, divide(1, 6))) | divide(n0,n3)|divide(n0,n5)|divide(n0,n1)|multiply(#0,#1)|subtract(n0,#2)|multiply(#3,#4)| | general |
how much is 75 % of 40 is greater than 4 / 5 of 25 ? | "( 75 / 100 ) * 40 β ( 4 / 5 ) * 25 = 10 answer : c" | a ) 22 , b ) 67 , c ) 10 , d ) 12 , e ) 66 | c | subtract(multiply(40, divide(75, const_100)), multiply(divide(4, 5), 25)) | divide(n0,const_100)|divide(n2,n3)|multiply(n1,#0)|multiply(n4,#1)|subtract(#2,#3)| | general |
a number is doubled and 5 is added . if the resultant is trebled , it becomes 105 . what is that number ? | "explanation : let the number be x . therefore , 3 ( 2 x + 5 ) = 105 6 x + 15 = 105 6 x = 90 x = 15 answer : d" | a ) 12 , b ) 29 , c ) 27 , d ) 15 , e ) 99 | d | divide(subtract(105, multiply(const_3, 5)), multiply(const_3, const_2)) | multiply(n0,const_3)|multiply(const_2,const_3)|subtract(n1,#0)|divide(#2,#1)| | general |
10 play kabadi , 35 play kho kho only , 5 play both gmaes . then how many in total ? | "10 play kabadi = > n ( a ) = 10 , 5 play both gmaes . = > n ( anb ) = 5 35 play kho kho only , = > n ( b ) = n ( b only ) + n ( anb ) = 35 + 5 = 40 total = > n ( aub ) = n ( a ) + n ( b ) - n ( anb ) = 10 + 40 - 5 = 45 answer : e" | a ) 30 , b ) 35 , c ) 38 , d ) 40 , e ) 45 | e | subtract(add(10, add(35, 5)), 5) | add(n1,n2)|add(n0,#0)|subtract(#1,n2)| | general |
if the average of 12 x and 8 y is greater than 200 , and x is twice y , what is the least integer value of x ? | substitution can be used in the following way : always start with the equation : x = 2 y . it is more straight forward to manage as compared to the inequality . substitute y = x / 2 , not the other way because you need to find the minimum value of x . so you can get rid of y . now go on to the inequality . so 8 y = 8 x... | a ) 20 , b ) 30 , c ) 40 , d ) 25 , e ) 35 | d | multiply(12, const_2) | multiply(n0,const_2) | general |
i sold a book at a profit of 10 % . had i sold it for $ 150 more , 15 % would have been gained . find the cost price ? | "115 % of cost - 110 % of cost = $ 150 5 % of cost = $ 150 cost = 150 * 100 / 5 = $ 3000 answer is c" | a ) $ 2000 , b ) $ 2500 , c ) $ 3000 , d ) $ 3120 , e ) $ 1540 | c | divide(multiply(divide(multiply(150, 10), subtract(15, 10)), const_100), 10) | multiply(n0,n1)|subtract(n2,n0)|divide(#0,#1)|multiply(#2,const_100)|divide(#3,n0)| | gain |
in a certain company , the ratio of the number of managers to the number of non - managers in any department must always be greater than 7 : 24 . in the company , what is the maximum number of non - managers in a department that has 8 managers ? | 8 / 7 * 24 = 27.4 the answer is b . | a ) 26 , b ) 27 , c ) 28 , d ) 29 , e ) 30 | b | floor(divide(multiply(24, 8), 7)) | multiply(n1,n2)|divide(#0,n0)|floor(#1) | general |
add 8 % of 24 and 10 % of 40 . | "8 % of 24 + 10 % of 40 24 * 8 / 100 + 40 * 10 / 100 1.9 + 4 = 5.9 answer d" | a ) 9.5 , b ) 10.5 , c ) 5.5 , d ) 5.9 , e ) 8.5 | d | add(divide(multiply(8, 24), const_100), divide(multiply(10, 40), const_100)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|add(#2,#3)| | gain |
in a certificate by mistake a candidate gave his height as 20 % more than actual height . in the interview panel , he clarified that his height was 5 feet 5 nches . find the % correction made by the candidate from his stated height to his actual height ? | "his height was = 5 feet 5 inch = 5 + 60 = 65 inch . required % correction = 65 * ( 1.20 - 1 ) = 13 a" | a ) 13 , b ) 20 , c ) 30 , d ) 43 , e ) 50 | a | multiply(divide(subtract(add(20, add(const_0_25, const_0_25)), 5), add(const_100, subtract(add(20, add(const_0_25, const_0_25)), 5))), const_100) | add(const_0_25,const_0_25)|add(n0,#0)|subtract(#1,n1)|add(#2,const_100)|divide(#2,#3)|multiply(#4,const_100)| | gain |
if 2 men or 3 women can reap a field in 12 days how long will 4 men and 5 women take to reap it ? | "explanation : 2 men reap 1 / 12 field in 1 day 1 man reap 1 / ( 2 x 12 ) 3 women reap 1 / 12 field in 1 day 1 woman reap 1 / ( 12 x 3 ) 4 men and 5 women reap ( 4 / ( 2 x 12 ) + 5 / ( 3 x 12 ) = 11 / 36 in 1 day 4 men and 5 women will reap the field in 11 / 36 days answer : option b" | a ) 12 / 16 , b ) 11 / 36 , c ) 14 / 24 , d ) 18 / 36 , e ) 2 / 24 | b | divide(12, add(multiply(4, divide(12, 2)), multiply(5, divide(12, 3)))) | divide(n2,n0)|divide(n2,n1)|multiply(n3,#0)|multiply(n4,#1)|add(#2,#3)|divide(n2,#4)| | physics |
one pack of cake mix is being stirred before being poured into two separate baking trays . the first tray can hold a certain amount of cups of the mix while the other holds 20 cups less than the first . if the total number of cups poured is 500 , how many cups of batter does the second tray hold ? | tray one as t 1 and tray two as t 2 . t 2 = t 1 - 20 cups as equation a . t 1 + t 2 = 500 cups in total as equation b . substitute equation a into equation b , resulting in t 1 + ( t 1 - 20 ) = 500 transpose like terms , 2 * t 1 = 500 + 20 ( 2 * t 1 ) / 2 = 520 / 2 t 1 = 260 , thus t 2 is t 1 less 20 cups , which is 24... | a ) 250 , b ) 260 , c ) 500 , d ) 280 , e ) 240 | e | divide(subtract(500, 20), const_2) | subtract(n1,n0)|divide(#0,const_2) | general |
a , b and c can do a work in 7 , 14 and 21 days respectively . they completed the work and got $ 242 . what is the share of c ? | "c $ 44 the ratio of their working rates = 1 / 7 : 1 / 14 : 1 / 21 = 6 : 3 : 2 . since , they work together , the share of c = 2 / 11 * 242 = $ 44" | a ) $ 25 , b ) $ 48 , c ) $ 44 , d ) $ 49 , e ) $ 40 | c | multiply(242, divide(inverse(14), add(inverse(21), add(inverse(7), inverse(14))))) | inverse(n1)|inverse(n0)|inverse(n2)|add(#1,#0)|add(#3,#2)|divide(#0,#4)|multiply(n3,#5)| | physics |
a certain no . when divided by 95 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ? | "explanation : 95 + 25 = 120 / 15 = 8 ( remainder ) d" | a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 9 | d | subtract(subtract(subtract(95, 25), const_4), const_2) | subtract(n0,n1)|subtract(#0,const_4)|subtract(#1,const_2)| | general |
during a sale of 20 % on everything in a store , a kid is successful in convincing the store manager to give him 20 candies for the discounted price of 14 candies . the store still makes a profit of 10 % on this sale . what is the mark up percentage on each candy ? | "hi akhil , i can share the way i solved it . . let ' s say marked price = m so , there ' s a discount of 20 % on m so , new s . p . = 80 % of m now , the child convinces the owner to sell 20 candies for the price of 14 candies . let ' s say each candy after discount is 1 $ . so , s . p . of 20 candies = 20 $ . the chi... | a ) 100 % , b ) 80 % , c ) 75 % , d ) 66 + 2 / 3 % , e ) 55 % | c | multiply(subtract(divide(add(divide(10, const_100), const_1), multiply(subtract(const_1, divide(20, const_100)), divide(subtract(20, 14), 20))), const_1), const_100) | divide(n3,const_100)|divide(n0,const_100)|subtract(n1,n2)|add(#0,const_1)|divide(#2,n1)|subtract(const_1,#1)|multiply(#4,#5)|divide(#3,#6)|subtract(#7,const_1)|multiply(#8,const_100)| | gain |
a and b are two multiples of 14 , and q is the set of consecutive integers between a and b , inclusive . if q contains 14 multiples of 14 , how many multiples of 7 are there in q ? | "halfway between the multiples of 14 , there will be another multiple of 7 . the total number of multiples of 7 is 14 + 13 = 27 . the answer is b ." | a ) 26 , b ) 27 , c ) 28 , d ) 29 , e ) 30 | b | subtract(multiply(14, const_2), const_1) | multiply(n1,const_2)|subtract(#0,const_1)| | physics |
the length of a train and that of a platform are equal . if with a speed of 180 k / hr , the train crosses the platform in one minute , then the length of the train ( in meters ) is ? | "speed = [ 180 * 5 / 18 ] m / sec = 50 m / sec ; time = 1 min . = 60 sec . let the length of the train and that of the platform be x meters . then , 2 x / 60 = 50 = > x = 50 * 60 / 2 = 1500 answer : a" | a ) 1500 , b ) 266 , c ) 299 , d ) 750 , e ) 261 | a | divide(divide(multiply(180, const_1000), divide(const_60, const_1)), const_2) | divide(const_60,const_1)|multiply(n0,const_1000)|divide(#1,#0)|divide(#2,const_2)| | physics |
running at the same constant rate , 8 identical machines can produce a total of 240 pens per minute . at this rate , how many pens could 5 such machines produce in 5 minutes ? | "let ' s take the approach that uses the answer choices to eliminate wasted time . 240 / 8 = 30 pens per minute per machine . 5 machines = 150 per minute . 5 minutes worth = 750 pens . looking at the answers it is clear . . . we can only choose ( d ) . the correct answer is d ." | a ) 114 , b ) 108 , c ) 192 , d ) 750 , e ) 777 | d | multiply(multiply(divide(240, 8), 5), 5) | divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)| | gain |
the product of the squares of two positive integers is 81 . how many pairs of positive integers satisfy this condition ? | "ans : c - 2 pairs ( x Λ 2 ) ( y Λ 2 ) = 81 [ square root both sides ] xy = 9 9 = 1 x 9 , 9 x 1 , 3 x 3 cancel the repeats this leaves us with exactly 2 options . hence , c" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | subtract(add(const_2, const_3), const_2) | add(const_2,const_3)|subtract(#0,const_2)| | geometry |
before 2 years , dog a β s age was 4 times of dog b β s age and after 4 years , dog a β s age will be 3 times of dog b β s age . what is the difference of dog a β s age and dog b β s now ? | a - 2 = 4 ( b - 2 ) - - > a - 4 b = - 6 . . . . . . . . . . . . . 1 a + 4 = 3 ( b + 4 ) - - > a - 3 b = 8 . . . . . . . . . . . . . 2 ( 2 ) - ( 1 ) - - > b = 14 - - > a = 3 ( 18 ) = 50 a - b = 50 - 14 = 36 answer : a | a ) 36 , b ) 42 , c ) 54 , d ) 60 , e ) 64 | a | subtract(subtract(multiply(4, add(subtract(multiply(4, 3), 4), subtract(multiply(4, 2), 2))), subtract(multiply(4, 2), 2)), add(subtract(multiply(4, 3), 4), subtract(multiply(4, 2), 2))) | multiply(n1,n2)|multiply(n0,n1)|subtract(#0,n1)|subtract(#1,n0)|add(#2,#3)|multiply(n1,#4)|subtract(#5,#3)|subtract(#6,#4)| | general |
how many integers between 500 and 1000 are there such that their unit digit is even ? | 500 numbers between - 500 and 1000 out of which half would be even , half odd . number of even unit digit number = 250 . correct option is a | a ) 250 , b ) 150 , c ) 500 , d ) 100 , e ) 200 | a | add(divide(500, const_2), const_1) | divide(n0,const_2)|add(#0,const_1) | general |
a man saves a certain portion of his income during a year and spends the remaining portion on his personal expenses . next year his income increases by 50 % but his savings increase by 100 % . if his total expenditure in 2 years is double his expenditure in 1 st year , what % age of his income in the first year did he ... | "1 st year income = i 1 st year savings = s 1 st year expense = e 1 2 nd year income = 1.5 i 2 nd year savings = 2 s ( 100 % increase ) 2 nd year expense = e 2 e 1 + e 2 = 2 e 1 e 2 = e 1 that means expenses are same during both years . with increase of 50 % income the savings increased by 100 % . or s = . 5 i or s = 5... | a ) 40 % , b ) 45 % , c ) 50 % , d ) 60 % , e ) 70 % | c | multiply(divide(subtract(add(add(100, 50), 100), multiply(2, 100)), 100), 100) | add(n0,n1)|multiply(n1,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,n1)|multiply(#4,n1)| | general |
the cost price of 24 articles is the same as the selling price of x articles . if the profit is 20 % , what is x ? | "let the cost price = y the cost price of 24 articles = 24 y the selling price of x articles = 1.20 y * x 1.20 y * x = 24 y x = 24 / 1.2 = 20 the answer is d ." | a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | d | divide(multiply(24, const_4), add(const_4, const_1)) | add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)| | gain |
on increasing the number of lines in a page by 110 , they become 240 . what is the % of increase in the no . of lines in the page ? | "explanation : number of pages increased = 110 now , the number of pages of book = 240 number of pages of the books before increase = 240 β 110 = 130 % increase in the number of pages in the book = 110 / 130 x 100 % = 84.6 % e" | a ) 20 % , b ) 305 , c ) 50 % , d ) 55 % , e ) 84.6 % | e | subtract(multiply(divide(240, subtract(240, 110)), const_100), const_100) | subtract(n1,n0)|divide(n1,#0)|multiply(#1,const_100)|subtract(#2,const_100)| | general |
an empty fuel tank with a capacity of 200 gallons was filled partially with fuel a and then to capacity with fuel b . fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume . if the full fuel tank contains 18 gallons of ethanol , how many gallons of fuel a were added ? | "say there are a gallons of fuel a in the tank , then there would be 200 - a gallons of fuel b . the amount of ethanol in a gallons of fuel a is 0.12 a ; the amount of ethanol in 200 - a gallons of fuel b is 0.16 ( 200 - a ) ; since the total amount of ethanol is 18 gallons then 0.12 a + 0.16 ( 200 - a ) = 18 - - > a =... | a ) 160 , b ) 150 , c ) 100 , d ) 80 , e ) 350 | e | divide(subtract(multiply(200, divide(16, const_100)), 18), subtract(divide(16, const_100), divide(12, const_100))) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|subtract(#0,#1)|subtract(#2,n3)|divide(#4,#3)| | gain |
a discount electronics store normally sells all merchandise at a discount of 10 percent to 40 percent off the suggested retail price . if , during a special sale , an additional 5 percent were to be deducted from the discount price , what would be the lowest possible price of an item costing $ 500 before any discount ? | "original price : 500 $ max first discount = - 40 % thus : 500 Γ’ Λ β ( 1 Γ’ Λ β 40 / 100 ) = 300 second discount on the discounted price = - 20 % thus : 300 Γ’ Λ β ( 1 Γ’ Λ β 5 / 100 ) = 285 answer b ." | a ) $ 130.00 , b ) $ 285.00 , c ) $ 200.00 , d ) $ 258.00 , e ) $ 210.00 | b | multiply(500, divide(add(40, 5), const_100)) | add(n1,n2)|divide(#0,const_100)|multiply(n3,#1)| | gain |
what is the total number of d integers between 100 and 200 that are divisible by 3 ? | yes there is a different way of arriving at that answer . . . . u can also use airthmetic progression to get the answer since the first term to be divisble by 3 is 102 . . take that as a . . the starting no and since 198 is the last digit to be divisible by 3 take that as n . . . since the difference is 3 take that as ... | a ) 33 , b ) 32 , c ) 31 , d ) 30 , e ) 29 | a | divide(add(subtract(multiply(floor(divide(200, 3)), 3), add(3, multiply(3, floor(divide(100, 3))))), 3), 3) | divide(n1,n2)|divide(n0,n2)|floor(#0)|floor(#1)|multiply(n2,#2)|multiply(n2,#3)|add(n2,#5)|subtract(#4,#6)|add(n2,#7)|divide(#8,n2) | general |
10 women can complete a work in 7 days and 10 children take 14 days to complete the work . how many days will 5 women and 10 children take to complete the work ? | explanation : 1 woman ' s 1 day ' s work = 1 / 70 1 child ' s 1 day ' s work = 1 / 140 5 women and 10 children 1 day work = ( 570 + 10140 ) = 17 so 5 women and 10 children will finish the work in 7 days answer : b | a ) 6 days , b ) 7 days , c ) 8 days , d ) 9 days , e ) none of these | b | inverse(add(divide(5, multiply(10, 7)), divide(10, multiply(10, 14)))) | multiply(n0,n1)|multiply(n0,n3)|divide(n4,#0)|divide(n0,#1)|add(#2,#3)|inverse(#4) | physics |
the average weight of a group of boys is 30 kg . after a boy of weight 38 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 30 x after the boy weighing 38 kg joins the group , total weight of boys = 30 x + 38 so 30 x + 38 = 31 ( x + 1 ) = > x = 7 . answer : e" | a ) a ) 4 , b ) b ) 8 , c ) c ) 6 , d ) d ) 2 , e ) e ) 7 | e | add(subtract(38, add(30, 1)), 1) | add(n0,n2)|subtract(n1,#0)|add(#1,n2)| | general |
the output of a factory is increased by 10 % to keep up with rising demand . to handle the holiday rush , this new output is increased by 20 % . by approximately what percent would the output of the factory now have to be decreased in order to restore the original output ? | "let the output of a factory is 100 first , increased by 10 % = 110 due to holiday rush , it increase by 20 % = 110 + 22 = 132 132 - ( x / 100 * 132 ) = 100 24.24 answer : b" | a ) 20 % , b ) 24 % , c ) 30 % , d ) 32 % , e ) 79 % | b | multiply(divide(subtract(multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 20), const_100)), const_1), multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 20), const_100))), const_100) | add(n0,const_100)|add(n1,const_100)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|subtract(#4,const_1)|divide(#5,#4)|multiply(#6,const_100)| | general |
maria sold 10 kg of apples in her first hour at the market , but only 2 kg of apples in the second hour . on average , how many kg of apples did she sell in two hours at the market ? | ( 10 kg + 2 kg ) / 2 = 6 kg correct answer is : a | a ) 6 kg , b ) 5 kg , c ) 12 kg , d ) 24 kg , e ) 20 kg | a | subtract(10, multiply(2, 2)) | multiply(n1,n1)|subtract(n0,#0) | physics |
a can finish a piece of work in 15 days . b can do it in 30 days . they work together for 6 days and then a goes away . in how many days will b finish the work ? | "6 / 15 + ( 6 + x ) / 30 = 1 = > x = 12 days answer : b" | a ) 11 days , b ) 12 days , c ) 14 days , d ) 10 days , e ) 15 days | b | divide(subtract(const_1, add(multiply(divide(const_1, const_4.0), const_2), multiply(divide(const_1, 30), const_2))), divide(const_1, 30)) | divide(const_1,const_4.0)|divide(const_1,n1)|multiply(#0,const_2)|multiply(#1,const_2)|add(#2,#3)|subtract(const_1,#4)|divide(#5,#1)| | physics |
find the largest 5 digit number which is exactly divisible by 88 ? | "largest 5 digit number is 99999 after doing 99999 Γ· 88 we get remainder 31 hence largest 5 digit number exactly divisible by 88 = 99999 - 31 = 99968 e" | a ) 98765 , b ) 98907 , c ) 99944 , d ) 99954 , e ) 99968 | e | multiply(add(add(add(add(multiply(const_100, const_100), multiply(const_100, const_10)), multiply(const_100, const_3)), multiply(5, const_10)), const_3), 88) | multiply(const_100,const_100)|multiply(const_10,const_100)|multiply(const_100,const_3)|multiply(n0,const_10)|add(#0,#1)|add(#4,#2)|add(#5,#3)|add(#6,const_3)|multiply(n1,#7)| | general |
a club wants to mix 15 pounds of candy worth $ 8.00 per pound with candy worth $ 5.00 per pound to reduce the cost of the mixture to $ 6.00 per pound . how many pounds of the $ 5.00 per pound candy should be used ? | "let number of pounds of 5 $ candy to be used be w 6 = ( 15 * 8 + 5 * w ) / ( 15 + w ) = > 90 + 6 w = 120 + 5 w = > w = 30 answer b" | a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | b | subtract(multiply(15, 8.00), multiply(6.00, 15)) | multiply(n0,n1)|multiply(n0,n3)|subtract(#0,#1)| | general |
300 + 5 Γ 8 = ? | 300 + 5 Γ 8 = ? or , ? = 300 + 40 = 340 answer b | a ) 820 , b ) 340 , c ) 420 , d ) 209 , e ) none of these | b | add(multiply(5, 8), 300) | multiply(n1,n2)|add(n0,#0) | general |
a trained covered x km at 40 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 3 x km . | "total time taken = x / 40 + 2 x / 20 hours = 5 x / 40 = x / 8 hours average speed = 3 x / ( x / 8 ) = 24 kmph answer : c" | a ) 18 kmph , b ) 17 kmph , c ) 24 kmph , d ) 19 kmph , e ) 12 kmph | c | divide(multiply(40, 3), add(divide(40, 40), divide(multiply(2, 40), 20))) | divide(n0,n0)|multiply(n0,n3)|multiply(n0,n1)|divide(#2,n2)|add(#0,#3)|divide(#1,#4)| | general |
the avg weight of a , b & c is 70 kg . if d joins the group , the avg weight of the group becomes 70 kg . if another man e who weights is 3 kg more than d replaces a , then the avgof b , c , d & e becomes 68 kg . what is the weight of a ? | "a + b + c = 3 * 70 = 210 a + b + c + d = 4 * 70 = 280 - - - - ( i ) so , d = 70 & e = 70 + 3 = 73 b + c + d + e = 68 * 4 = 272 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = 280 β 272 = 8 a = e + 8 = 73 + 8 = 81 answer : a" | a ) 81 , b ) 65 , c ) 75 , d ) 89 , e ) 90 | a | subtract(multiply(70, const_4), subtract(multiply(68, const_4), add(3, subtract(multiply(70, const_4), multiply(70, 3))))) | multiply(n1,const_4)|multiply(n3,const_4)|multiply(n0,n2)|subtract(#0,#2)|add(n2,#3)|subtract(#1,#4)|subtract(#0,#5)| | general |
john paid a sum of money for purchasing 30 pens , which he recovered in full when he sold 20 of them . what was his percentage of profit or loss per pen ? | "a 50 % if the sum he paid whilst purchasing 30 pens = a , then the cost price of each pen = a / 30 . since the amount he got whilst selling 20 pens is also = a then the selling price of each pen = a / 20 . since selling price > cost price , he made a profit . profit per pen = selling price - cost price = a / 20 - a / ... | a ) 50 % , b ) 60 % , c ) 40 % , d ) 30 % , e ) 45 % | a | multiply(divide(subtract(30, 20), 20), const_100) | subtract(n0,n1)|divide(#0,n1)|multiply(#1,const_100)| | gain |
1 / 0.025 is equal to | "explanation : 1 / 0.025 = ( 1 * 1000 ) / 25 = 1000 / 25 = 40 option c" | a ) 25.5 , b ) 2.5 , c ) 40 , d ) . 25 , e ) none of these | c | divide(1, 0.025) | divide(n0,n1)| | general |
find the circumference and area of radius 4 cm . | "area of circle = Ο r Β² = 22 / 7 Γ 4 Γ 4 cm Β² = 50 cm Β² answer : c" | a ) 24 cm Β² , b ) 44 cm Β² , c ) 50 cm Β² , d ) 84 cm Β² , e ) 94 cm Β² | c | circle_area(4) | circle_area(n0)| | geometry |
a train covers a distance of 18 km in 3 min . if it takes 4 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 18 / 3 * 60 ) km / hr = ( 360 * 5 / 18 ) m / sec = 100 m / sec . length of the train = 100 * 4 = 400 m . answer : option d" | a ) 370 , b ) 380 , c ) 390 , d ) 400 , e ) 410 | d | divide(18, subtract(divide(18, 3), 4)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics |
a merchant sells an item at a 20 % discount , but still makes a gross profit of 25 percent of the cost . what percent of the cost would the gross profit on the item have been if it had been sold without the discount ? | "original sp = x cost = c current selling price = . 8 x ( 20 % discount ) . 8 x = 1.5 c ( 25 % profit ) x = 1.25 / . 8 * c x = 12.5 / 8 c original selling price is 1.5625 c which is 56.25 % profit answer c" | a ) 20 % , b ) 40 % , c ) 56.25 % , d ) 60 % , e ) 75 % | c | subtract(const_100, subtract(subtract(const_100, 20), 25)) | subtract(const_100,n0)|subtract(#0,n1)|subtract(const_100,#1)| | gain |
evaluate : | 4 - 8 ( 3 - 12 ) | - | 5 - 11 | = | "according to order of operations , inner brackets first . hence | 4 - 8 ( 3 - 12 ) | - | 5 - 11 | = | 4 - 8 * ( - 9 ) | - | 5 - 11 | according to order of operations , multiplication within absolute value signs ( which may be considered as brackets when it comes to order of operations ) next . hence = | 4 + 72 | - | 5... | a ) 49 , b ) 55 , c ) 69 , d ) 70 , e ) 82 | d | subtract(subtract(4, multiply(8, subtract(3, 12))), negate(subtract(5, 11))) | subtract(n2,n3)|subtract(n4,n5)|multiply(n1,#0)|negate(#1)|subtract(n0,#2)|subtract(#4,#3)| | general |
how much time will a train of length 200 m moving at a speed of 72 kmph take to cross another train of length 300 m , moving at 36 kmph in the same direction ? | the distance to be covered = sum of their lengths = 200 + 300 = 500 m . relative speed = 72 - 36 = 36 kmph = 36 * 5 / 18 = 10 mps . time required = d / s = 500 / 10 = 50 sec . answer : a | a ) 50 sec , b ) 26 sec , c ) 80 sec , d ) 82 sec , e ) 81 sec | a | divide(add(200, 300), multiply(subtract(72, 36), const_0_2778)) | add(n0,n2)|subtract(n1,n3)|multiply(#1,const_0_2778)|divide(#0,#2) | physics |
a train covers a distance of 36 km in 12 min . if it takes 7 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 36 / 12 * 60 ) km / hr = ( 180 * 5 / 18 ) m / sec = 50 m / sec . length of the train = 50 * 7 = 350 m . answer : option c" | a ) 330 , b ) 340 , c ) 350 , d ) 360 , e ) 370 | c | divide(36, subtract(divide(36, 12), 7)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics |
a sum of money deposited at c . i . amounts to rs . 2420 in 2 years and to rs . 3146 in 3 years . find the rate percent ? | "explanation : 2420 - - - 726 100 - - - ? = > 30 % answer : option b" | a ) 11 , b ) 30 , c ) 28 , d ) 24 , e ) 82 | b | multiply(divide(subtract(3146, 2420), 2420), const_100) | subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
the rate of spin of a certain gyroscope doubled every 10 seconds from the moment a particular stopwatch started . if after a minute and a half the gyroscope reached a speed of 800 meters per second , what was the speed , in meters per second , when the stopwatch was started ? | let x be the original speed when the stopwatch was started . in 90 seconds , the speed doubled 9 times . 2 ^ 9 * x = 800 x = ( 2 ^ 5 * 25 ) / 2 ^ 9 = 25 / 16 the answer is d . | a ) 25 / 3 , b ) 25 / 4 , c ) 25 / 8 , d ) 25 / 16 , e ) 25 / 32 | d | divide(divide(800, power(const_2, subtract(divide(add(divide(const_60, const_2), const_60), 10), const_1))), const_2) | divide(const_60,const_2)|add(#0,const_60)|divide(#1,n0)|subtract(#2,const_1)|power(const_2,#3)|divide(n1,#4)|divide(#5,const_2) | physics |
if i equals the sum of the even integers from 2 to 224 , inclusive , and k equals the sum of the even integers from 8 to 80 , inclusive , what is the value of i - k ? | use following formulae for such problems : sum of evenly spaced integers = ( # of integers ) * ( mean of integers ) # of integers = [ ( last - first ) / 2 ] + 1 mean of integers = ( last + first ) / 2 in above problem : # of integers = [ ( 224 - 2 ) / 2 ] + 1 = 112 and [ ( 80 - 8 ) / 2 ] + 1 = 37 mean of integers = ( 2... | a ) 11028 , b ) 14172 , c ) 14284 , d ) 14015 , e ) 14397 | a | add(divide(add(multiply(add(add(80, 2), 224), divide(subtract(224, add(80, 2)), const_2)), add(add(80, 2), 224)), const_2), add(add(2, 2), 8)) | add(n0,n0)|add(n0,n3)|add(n2,#0)|add(n1,#1)|subtract(n1,#1)|divide(#4,const_2)|multiply(#3,#5)|add(#3,#6)|divide(#7,const_2)|add(#2,#8) | general |
if 25 % of x is 15 less than 12 % of 1500 , then x is ? | "25 % of x = x / 4 ; 12 % of 1500 = 12 / 100 * 1500 = 180 given that , x / 4 = 180 - 15 = > x / 4 = 165 = > x = 660 . answer : a" | a ) 660 , b ) 738 , c ) 837 , d ) 840 , e ) 83 | a | divide(subtract(multiply(1500, divide(12, const_100)), 15), divide(25, const_100)) | divide(n2,const_100)|divide(n0,const_100)|multiply(n3,#0)|subtract(#2,n1)|divide(#3,#1)| | general |
evaluate : 11 + sqrt ( - 4 + 6 Γ 4 Γ· 3 ) = . . . . | according to order of operations , inner brackets first where 6 Γ 4 Γ· 3 is first calculated since it has a multiplication and a division . 6 Γ 4 Γ· 3 = 24 Γ· 3 = 8 hence 11 + sqrt ( - 4 + 6 Γ 4 Γ· 3 ) = 11 + sqrt ( - 4 + 8 ) = 11 + sqrt ( 4 ) = 11 + 2 = 13 correct answer is e ) 13 | a ) 3 , b ) 33 , c ) 43 , d ) 23 , e ) 13 | e | add(11, sqrt(subtract(divide(multiply(6, 4), 3), 4))) | multiply(n1,n2)|divide(#0,n4)|subtract(#1,n1)|sqrt(#2)|add(n0,#3) | general |
a β palindromic integer β is an integer that remains the same when its digits are reversed . so , for example , 43334 and 516615 are both examples of palindromic integers . how many 6 - digit palindromic integers are both even and greater than 800,000 ? | "the first digit and last digit are the same so the only possibility is 8 . the second and third digits can be any number from 0 to 9 . the total number of palindromic integers is 1 * 10 * 10 = 100 the answer is b ." | a ) 60 , b ) 100 , c ) 160 , d ) 240 , e ) 300 | b | multiply(multiply(subtract(6, const_4), const_10), const_10) | subtract(n2,const_4)|multiply(#0,const_10)|multiply(#1,const_10)| | general |
a trader sold an item at a loss of 18 % . if he had increased the price by $ 25 he would have made a gain of 5 % . what is the cost price of the item ? | let c . p . be $ x then 105 % of x - 82 % of x = 100 23 % of x = 100 23 / 100 * x = 100 23 x = 10000 x = 434.78 answer is d | a ) $ 414.78 , b ) $ 424.78 , c ) $ 534.78 , d ) $ 434.78 , e ) $ 435.78 | d | divide(multiply(25, const_100), multiply(subtract(add(const_1, divide(5, const_100)), subtract(const_1, divide(18, const_100))), const_100)) | divide(n2,const_100)|divide(n0,const_100)|multiply(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|subtract(#3,#4)|multiply(#5,const_100)|divide(#2,#6) | gain |
if 3 women can color 180 m long cloth in 2 days , then 5 women can color 300 m long cloth in ? | "the length of cloth painted by one woman in one day = 180 / 3 Γ 2 = 30 m no . of days required to paint 300 m cloth by 5 women = 300 / 5 Γ 30 = 2 days answer : a" | a ) 2 days , b ) 3 days , c ) 5 days , d ) 1 day , e ) 4 days | a | divide(300, multiply(5, divide(180, multiply(3, 2)))) | multiply(n0,n2)|divide(n1,#0)|multiply(n3,#1)|divide(n4,#2)| | physics |
if $ 5,000 is invested in an account that earns 6 % interest compounded semi - annually , then the interest earned after one year would be how much greater than if the $ 5,000 had been invested at 8 % simple yearly interest ? | "ans a solution amount ( ci ) = p + ( 1 + r / n ) ^ nt = 5000 + ( 1 + 0.06 / 2 ) ^ 2 = 5406 amount ( si ) = p + ptr / 100 = 5000 + ( 5000 * 1 * 8 / 100 ) = 5400 difference = 5406 - 5400 = 6 $ a" | a ) $ 6 , b ) $ 8 , c ) $ 12 , d ) $ 16 , e ) $ 432 | a | multiply(6, const_1) | multiply(n1,const_1)| | gain |
a train with 1200 m crosses a tree in 120 sec , how much time will i take to pass a platform 700 m long ? | s = 1200 / 120 s - 10 m / sec total length = 1900 m t = d / s t = 1900 / 10 t = 190 sec answer c | a ) 150 sec , b ) 130 sec , c ) 190 sec , d ) 200 sec , e ) 100 sec | c | divide(add(1200, 700), divide(1200, 120)) | add(n0,n2)|divide(n0,n1)|divide(#0,#1) | physics |
the area of a circle is added to its diameter , and the circumference is then subtracted from this total , the result is 10 . what is the radius of the circle ? | the equation is ; diameter + area - circumference = d + a - c = d + pi * r ^ 2 - 2 * pi * r = 2 r + pi * r ( r - 2 ) . by plugging in the answers we can test the answers quickly ; then , 4 is the only possible answer . answer : e | ['a ) 2 * pi', 'b ) pi', 'c ) 2', 'd ) 4 * pi', 'e ) 4'] | e | sqrt(subtract(power(divide(multiply(subtract(const_1, const_pi), const_2), const_pi), const_2), multiply(const_4, negate(divide(10, const_pi))))) | divide(n0,const_pi)|subtract(const_1,const_pi)|multiply(#1,const_2)|negate(#0)|divide(#2,const_pi)|multiply(#3,const_4)|power(#4,const_2)|subtract(#6,#5)|sqrt(#7) | geometry |
a circular well with a diameter of 2 metres , is dug to a depth of 8 metres . what is the volume of the earth dug out ? | "solution volume = Ο r 2 h βΉ = βΊ ( 22 / 7 Γ 1 Γ 1 Γ 8 ) m 3 βΉ = βΊ 25.1 m 3 . answer b" | a ) 32 m 3 , b ) 25.1 m 3 , c ) 40 m 3 , d ) 44 m 3 , e ) none | b | volume_cylinder(divide(2, 2), 8) | divide(n0,n0)|volume_cylinder(#0,n1)| | geometry |
a man can row upstream at 5 kmph and downstream at 25 kmph , and then find the speed of the man in still water ? | "us = 5 ds = 25 m = ( 5 + 25 ) / 2 = 15 answer : a" | a ) 15 , b ) 77 , c ) 30 , d ) 88 , e ) 34 | a | divide(add(5, 25), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
sum of 49 odd numbers is ? | sum of 1 st n odd no . s = 1 + 3 + 5 + 7 + . . . = n ^ 2 so , sum of 1 st 49 odd numbers = 49 ^ 2 = 2401 answer : d | a ) 2398 , b ) 2399 , c ) 2400 , d ) 2401 , e ) 2402 | d | multiply(multiply(49, const_2), divide(49, const_2)) | divide(n0,const_2)|multiply(n0,const_2)|multiply(#0,#1) | general |
a walks at 30 kmph and 30 hours after his start , b cycles after him at 40 kmph . how far from the start does b catch up with a ? | "suppose after x km from the start b catches up with a . then , the difference in the time taken by a to cover x km and that taken by b to cover x km is 30 hours . x / 30 - x / 40 = 30 x = 3600 km answer is d" | a ) 1200 km , b ) 1500 km , c ) 2000 km , d ) 2500 km , e ) 3600 km | d | multiply(30, 40) | multiply(n1,n2)| | physics |
one side of a rectangular field is 16 m and one of its diagonal is 18 m . find the area of the field . | "solution other side = β ( 18 ) 2 - ( 16 ) 2 = β 324 - 256 = β 68 = 8.2 m . β΄ area = ( 16 x 8.2 ) m 2 = 131.2 m 2 . answer e" | a ) 100 , b ) 120 , c ) 150 , d ) 180 , e ) none | e | rectangle_area(16, sqrt(subtract(power(18, const_2), power(16, const_2)))) | power(n1,const_2)|power(n0,const_2)|subtract(#0,#1)|sqrt(#2)|rectangle_area(n0,#3)| | geometry |
the product of x and y is a constant . if the value of x is increased by 20 % , by what percentage must the value of y be decreased ? | "x * y = constt . let x = y = 100 in beginning i . e . x * y = 100 * 100 = 10000 x ( 100 ) - - - becomes - - - > 1.2 x ( 120 ) i . e . 120 * new ' y ' = 10000 i . e . new ' y ' = 10000 / 120 = 83.33 i . e . y decreases from 100 to 83.33 i . e . decrease of 16.66 % answer : option a" | a ) 16.66 % , b ) 33.33 % , c ) 44.44 % , d ) 55.55 % , e ) 19.92 % | a | multiply(subtract(const_1, divide(const_100, add(const_100, 20))), const_100) | add(n0,const_100)|divide(const_100,#0)|subtract(const_1,#1)|multiply(#2,const_100)| | general |
if 12 men or 20 women can do a piece of work in 27 days , then in how many days can 9 men and 12 women together do the work ? | "c 20 days given that 12 m = 20 w = > 3 m = 5 w 9 men + 12 women = 15 women + 12 women = 27 women 20 women can do the work in 27 days . so , 27 women can do it in ( 20 * 27 ) / 27 = 20 days ." | a ) 10 days , b ) 30 days , c ) 20 days , d ) 80 days , e ) 40 days | c | inverse(add(divide(9, multiply(12, 27)), divide(12, multiply(20, 27)))) | multiply(n0,n2)|multiply(n1,n2)|divide(n3,#0)|divide(n4,#1)|add(#2,#3)|inverse(#4)| | physics |
how many trucks are there if each truck carrying 70 packages and total of 490 packages ? | sol . total packages 490 each truck carries 70 packages = 490 / 70 = 7 answer : a | a ) a ) 7 , b ) b ) 6 , c ) c ) 9 , d ) d ) 11 , e ) e ) none of the above | a | divide(490, 70) | divide(n1,n0) | general |
calculate 42120 Γ· ? = 195 | "answer let 42120 / x = 195 then x = 42120 / 195 = 216 option : e" | a ) 212 , b ) 219 , c ) 217 , d ) 213 , e ) 216 | e | multiply(42120, 195) | multiply(n0,n1)| | general |
the average of 6 observations is 14 . a new observation is included and the new average is decreased by 1 . the seventh observation is ? | "let seventh observation = x . then , according to the question we have = > ( 84 + x ) / 7 = 13 = > x = 7 . hence , the seventh observation is 7 . answer : e" | a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 7 | e | subtract(multiply(subtract(14, 1), add(6, 1)), multiply(14, 6)) | add(n0,n2)|multiply(n0,n1)|subtract(n1,n2)|multiply(#0,#2)|subtract(#3,#1)| | general |
find the value of ( 50 + 20 / 90 ) Γ 90 | "( 4500 + 20 ) / 90 * 90 = 4520 answer : c" | a ) 2520 , b ) 3520 , c ) 4520 , d ) 1520 , e ) 1750 | c | multiply(add(divide(20, 90), 50), 90) | divide(n1,n2)|add(n0,#0)|multiply(#1,n2)| | general |
if a and b are positive real numbers , and a ^ 3 + b ^ 3 = 20 , then the greatest possible value of a is between : | "if a > 3 , then a ^ 3 + b ^ 3 > 20 . the answer is a ." | a ) 0 and 3 , b ) 3 and 4 , c ) 4 and 5 , d ) 5 and 7 , e ) 7 and 9 | a | power(20, divide(const_1, 3)) | divide(const_1,n0)|power(n2,#0)| | general |
the ages of 2 persons differ by 32 years . if 9 years ago the elder one be 5 times as old as the younger one , find the present age of elder person . | "age of the younger person = x age of the elder person = x + 32 5 ( x - 9 ) = x + 32 - 9 x = 17 age of elder person = 17 + 32 = 49 answer is d" | a ) 30 , b ) 42.5 , c ) 45 , d ) 49 , e ) 47.9 | d | subtract(add(divide(multiply(2, 32), subtract(32, const_1)), 32), 2) | multiply(n0,n1)|subtract(n1,const_1)|divide(#0,#1)|add(n1,#2)|subtract(#3,n0)| | general |
what is the greatest possible length which can be used to measure exactly the lengths 18 m , 40 m and 6 m 50 cm ? | "required length = hcf of 1800 cm , 4000 cm , 650 cm = 50 cm answer is d" | a ) 20 cm , b ) 23 cm , c ) 25 cm , d ) 10 cm , e ) 28 cm | d | multiply(50, const_4) | multiply(n3,const_4)| | physics |
from january 1 , 2015 , to january 1 , 2017 , the number of people enrolled in health maintenance organizations increased by 5 percent . the enrollment on january 1 , 2017 , was 45 million . how many million people , to the nearest million , were enrolled in health maintenance organizations on january 1 , 2015 ? | "soln : - 5 x = 45 - - > 21 / 20 * x = 45 - - > x = 45 * 20 / 21 = 300 / 7 = ~ 43 . answer : e ." | a ) 38 , b ) 39 , c ) 40 , d ) 41 , e ) 43 | e | multiply(divide(const_100, add(const_100, 5)), 45) | add(n4,const_100)|divide(const_100,#0)|multiply(n7,#1)| | gain |
the number which exceeds 16 % of it by 42 is : | "solution solution let the number be x . x - 16 % of x = 42 x - 16 / 100 x = 42 x - 4 / 25 x = 42 21 / 25 x = 42 x = ( 42 x 25 / 21 ) = 50 answer a" | a ) 50 , b ) 52 , c ) 58 , d ) 60 , e ) 62 | a | divide(multiply(42, const_100), subtract(const_100, 16)) | multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain |
in assembling a bluetooth device , a factory uses one of two kinds of modules . one module costs $ 10 and the other one , that is cheaper , costs $ 2.5 . the factory holds a $ 62.50 worth stock of 22 modules . how many of the modules in the stock are of the cheaper kind ? | "so the number of $ 2.50 modules must be 21 so that the leftover 1 modules are of $ 10 which will give a total value $ 62.50 . 21 * 2.50 + 1 * 10 = 52.50 + 10 = 62.50 answer : c" | a ) 31 , b ) 35 , c ) 21 , d ) 40 , e ) 45 | c | divide(subtract(62.50, 10), 2.5) | subtract(n2,n0)|divide(#0,n1)| | other |
mary goes into labor at her local grocery store and is rushed to a hospital in an ambulance traveling 60 mph . her husband don drives after the ambulance at an average speed of 30 mph . mary reaches the hospital fifteen minutes later . how long does it take don to get there from the store ? | distance covered by the ambulance in 15 minutes = 15 miles 60 mph is one mile per minute , therefore 30 mph will be 1 / 2 a mile per minute . so , he reaches the hospital half an hour after leaving the store . correct answer : b | a ) 25 minutes , b ) 30 minutes , c ) 10 minutes , d ) 15 minutes , e ) 7 minutes | b | add(add(const_3, const_12), multiply(divide(multiply(30, divide(add(const_3, const_12), const_60)), subtract(60, 30)), const_60)) | add(const_12,const_3)|subtract(n0,n1)|divide(#0,const_60)|multiply(n1,#2)|divide(#3,#1)|multiply(#4,const_60)|add(#0,#5) | physics |
what is the ratio between perimeters of two squares one having 7 times the diagonal then the other ? | "d = 7 d d = d a β 2 = 7 d a β 2 = d a = 7 d / β 2 a = d / β 2 = > 7 : 1 answer : d" | a ) 3 : 7 , b ) 3 : 5 , c ) 3 : 8 , d ) 7 : 1 , e ) 3 : 5 | d | divide(7, divide(7, 7)) | divide(n0,n0)|divide(n0,#0)| | geometry |
square a has an area of 65 square centimeters . square b has a perimeter of 16 centimeters . if square b is placed within square a and a random point is chosen within square a , what is the probability the point is not within square b ? | "i guess it ' s mean that square b is placed within square aentirely . since , the perimeter of b is 16 , then its side is 16 / 4 = 4 and the area is 4 ^ 2 = 16 ; empty space between the squares is 65 - 16 = 48 square centimeters , so if a random point is in this area then it wo n ' t be within square b : p = favorable... | a ) 0.25 , b ) 0.5 , c ) 0.75 , d ) 0.1 , e ) 0.3 | c | divide(subtract(65, power(divide(16, const_4), const_2)), 65) | divide(n1,const_4)|power(#0,const_2)|subtract(n0,#1)|divide(#2,n0)| | geometry |
in an election , candidate a got 80 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favor of candidate . | "total number of invalid votes = 15 % of 560000 = 15 / 100 Γ 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 β 84000 = 476000 percentage of votes polled in favour of candidate a = 80 % therefore , the number of valid votes polled in favour of candidate a = 80 % of 476000 = 80 / 100 Γ 476000 = 38080000... | a ) 330000 , b ) 340000 , c ) 380800 , d ) 356000 , e ) 357000 | c | multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(80, const_100)) | divide(n0,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(n2,#2)|multiply(#0,#3)| | gain |
a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 560 sq . feet , how many feet of fencing will be required ? | "given that length and area , so we can find the breadth . length x breadth = area 20 x breadth = 560 breadth = 28 feet area to be fenced = 2 b + l = 2 ( 28 ) + 20 = 76 feet answer : d" | a ) 34 , b ) 40 , c ) 68 , d ) 76 , e ) 92 | d | add(multiply(divide(560, 20), const_2), 20) | divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)| | geometry |
if a man earns rs 20 on first day and spends rs 15 on second day , and earns rs 20 on third day and spends rs 15 on fourth day and goes on on which day he will be having rs 60 . | earns rs 20 on first day and spends rs 15 on second day means , end of day 2 , he have 5 rs . therefore , in 16 days he have 40 rs 2 * 8 = 16 days 5 rs * 8 = 40 rs and on the 17 th day again he earn rs . 20 , 40 + 20 = 60 rs . answer : e | a ) 40 rs , b ) 45 rs , c ) 50 rs , d ) 55 rs , e ) 60 rs | e | add(multiply(divide(add(const_3, const_2), const_2), multiply(const_4, const_4)), 20) | add(const_2,const_3)|multiply(const_4,const_4)|divide(#0,const_2)|multiply(#2,#1)|add(n0,#3) | physics |
4 mat - weaves can weaves 4 mats in 4 days . at the same rate , how many mats would be woven by 8 mat - weaves in 8 days ? | solution let the required number of mats be x . more weaves , more mats ( direct proportion ) more days , more mats ( direct proportion ) β΄ 4 Γ 4 Γ x = 8 Γ 8 Γ 4 β x = ( 8 x 8 x 4 ) / ( 4 x 4 ) = 16 . answer d | a ) 4 , b ) 8 , c ) 12 , d ) 16 , e ) 17 | d | divide(multiply(multiply(4, 8), 8), multiply(4, 4)) | multiply(n0,n3)|multiply(n0,n0)|multiply(n3,#0)|divide(#2,#1) | gain |
the average age of 36 students in a group is 14 years . when teacher β s age is included to it , the average increases by one . what is the teacher β s age in years ? | "sol . age of the teacher = ( 37 Γ 15 β 36 Γ 14 ) years = 51 years . answer d" | a ) 31 , b ) 36 , c ) 41 , d ) 51 , e ) none | d | add(36, const_1) | add(n0,const_1)| | general |
two trains of equal length are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train catches and completely passes the slower train in 18 seconds . what is the length of each train ( in meters ) ? | "the relative speed = 46 - 36 = 10 km / hr = 10 * 5 / 18 = 25 / 9 m / s in 18 seconds , the relative difference in distance traveled is 18 * 25 / 9 = 50 meters this distance is twice the length of each train . the length of each train is 50 / 2 = 25 meters the answer is b ." | a ) 20 , b ) 25 , c ) 35 , d ) 50 , e ) 65 | b | divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 18), const_2) | subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)| | physics |
the numbers 272738 and 232342 , when divided by n , a 2 digit number leave a remainder 13 and 17 respectively . find the sum of digits of n ? | "we have to find the h . c . f of both 1.272738 - 13 2.232325 - 17 h . c . f is 25 ans is 7 answer : d" | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | add(reminder(gcd(subtract(272738, 2), subtract(232342, 13)), const_10), const_2) | subtract(n0,n2)|subtract(n1,n3)|gcd(#0,#1)|reminder(#2,const_10)|add(#3,const_2)| | general |
a train is 360 meter long is running at a speed of 90 km / hour . in what time will it pass a bridge of 140 meter length ? | "speed = 90 km / hr = 90 * ( 5 / 18 ) m / sec = 25 m / sec total distance = 360 + 140 = 500 meter time = distance / speed = 500 * ( 1 / 25 ) = 20 seconds answer : e" | a ) 27 seconds , b ) 29 seconds , c ) 40 seconds , d ) 11 seconds , e ) 20 seconds | e | divide(add(360, 140), divide(multiply(90, const_1000), const_3600)) | add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)| | physics |
for each month of a given year except december , a worker earned the same monthly salary and donated one - tenth of that salary to charity . in december , the worker earned n times his usual monthly salary and donated one - fourth of his earnings to charity . if the worker ' s charitable contributions totaled one - eig... | "let monthly salary for each of the 11 months except december was x , then 11 x * 1 / 10 + nx * 1 / 4 = 1 / 8 ( 11 x + nx ) ; 11 / 10 + n / 4 = 1 / 8 ( 11 + n ) = > 44 + 10 n / 40 = 11 + n / 8 352 + 80 n = 440 + 40 n = > 40 n = 88 n = 88 / 40 = 22 / 10 answer : d ." | a ) 8 / 5 , b ) 5 / 2 , c ) 3 , d ) 22 / 10 , e ) 4 | d | divide(multiply(subtract(const_12, const_1), subtract(inverse(subtract(const_12, const_3)), inverse(const_10))), subtract(inverse(add(const_1, const_4)), inverse(subtract(const_12, const_3)))) | add(const_1,const_4)|inverse(const_10)|subtract(const_12,const_1)|subtract(const_12,const_3)|inverse(#3)|inverse(#0)|subtract(#4,#1)|subtract(#5,#4)|multiply(#2,#6)|divide(#8,#7)| | general |
if sharon ' s weekly salary increased by 16 percent , she would earn $ 348 per week . if instead , her weekly salary were to increase by 20 percent , how much would she earn per week ? | "( 348 / 116 ) 120 = 360 in this case long division does not take much time . ( 348 / 116 ) = 3 3 * 120 = 360 ( 300 + 60 ) answer d" | a ) $ 374 , b ) $ 382 , c ) $ 385 , d ) $ 360 , e ) $ 399 | d | add(divide(348, add(const_1, divide(16, const_100))), multiply(divide(20, const_100), divide(348, add(const_1, divide(16, const_100))))) | divide(n0,const_100)|divide(n2,const_100)|add(#0,const_1)|divide(n1,#2)|multiply(#1,#3)|add(#3,#4)| | general |
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