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 rectangular axis system , what is the area of a parallelogram with the coordinates : ( 1,3 ) , ( 5,3 ) , ( 2,6 ) , ( 6,6 ) ? | "delta x will give us the dimension of one side of the parallelogram = 5 - 1 = 4 unit delta y will give us the dimension of the other side of parallelogram = 6 - 3 = 3 unit area of parallelogram = 4 * 3 = 12 answer is b" | a ) 21 . , b ) 12 . , c ) 35 . , d ) 49 . , e ) 52 . | b | add(const_3, const_2) | add(const_2,const_3)| | geometry |
the banker ' s gain on a sum due 3 years hence at 12 % per year is rs . 270 . the banker ' s discount is : | "t . d . = ( b . g . x 100 / r x t ) = rs . ( 270 x 100 / 12 x 3 ) = rs . 750 . therefore , b . d . = rs . ( 750 + 270 ) = rs . 1020 . answer is c" | a ) 960 , b ) 840 , c ) 1020 , d ) 760 , e ) 920 | c | add(270, divide(multiply(270, const_100), multiply(12, 3))) | multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)|add(n2,#2)| | gain |
convert the 12 / 43 m / s into kilometers per hour ? | "12 / 43 m / s = 12 / 43 * 18 / 5 = 1 ( 1 / 250 ) = 1 kmph . answer : c" | a ) 3.5 kmph . , b ) 2.5 kmph . , c ) 1 kmph . , d ) 1.5 kmph . , e ) 1.9 kmph . | c | multiply(const_3_6, divide(12, 43)) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
a rectangular lawn of dimensions 80 m * 60 m has two roads each 10 m wide running in the middle of the lawn , one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . 4 per sq m ? | "area = ( l + b â € “ d ) d ( 80 + 60 â € “ 10 ) 10 = > 1300 m 2 1300 * 4 = rs . 5200 answer : d" | a ) 2288 , b ) 2779 , c ) 2779 , d ) 5200 , e ) 2781 | d | multiply(multiply(subtract(add(80, 60), 10), 10), 4) | add(n0,n1)|subtract(#0,n2)|multiply(n2,#1)|multiply(n3,#2)| | geometry |
a 45 ° - 45 ° - 90 ° right triangle has hypotenuse of length h . what is the area of the triangle r in terms of h ? | if . . . each of the two shorter sides = 3 , then the hypotenuse = h = 3 ( root 2 ) . the area r = ( 1 / 2 ) ( base ) ( height ) = ( 1 / 2 ) ( 3 ) ( 3 ) = 9 / 2 . so we ' re looking for an answer that = 9 / 2 when h = 3 ( root 2 ) . there ' s only one answer that matches . . . e | ['a ) h / √ 2', 'b ) h / 2', 'c ) h / 4', 'd ) ( h ) ^ 2', 'e ) ( h ) ^ 2 / 4'] | e | inverse(sine(45)) | sine(n0)|inverse(#0) | geometry |
- 88 * 49 + 100 = ? | "= > - 88 * ( 50 - 1 ) + 100 ; = > - ( 88 * 50 ) + 88 + 100 ; = > - 4400 + 188 = - 4212 . correct option : a" | a ) - 4212 , b ) 4601 , c ) - 4801 , d ) - 3471 , e ) none of these | a | add(multiply(negate(88), 49), 100) | negate(n0)|multiply(n1,#0)|add(n2,#1)| | general |
how many digits 2 ^ 200 has ? | "2 ^ 10 = 1.024 * 10 ^ 3 = > 2 ^ 100 = ( 1.024 ) ^ 10 * 10 ^ 60 therefore 61 digits would be my best guess b" | a ) 31 , b ) 61 , c ) 50 , d ) 99 , e ) 101 | b | floor(add(const_1, multiply(divide(log(2), log(const_10)), 200))) | log(n0)|log(const_10)|divide(#0,#1)|multiply(n1,#2)|add(#3,const_1)|floor(#4)| | general |
in an intercollegiate competition that lasted for 3 days , 175 students took part on day 1 , 210 on day 2 and 150 on day 3 . if 80 took part on day 1 and day 2 and 70 took part on day 2 and day 3 and 20 took part on all three days , how many students took part only on day 1 ? | day 1 & 2 = 80 ; only day 1 & 2 ( 80 - 20 ) = 60 , day 2 & 3 = 70 ; only day 2 & 3 ( 70 - 20 ) = 50 , only day 1 = 175 - ( 60 + 50 + 20 ) = 45 answer : b | a ) 25 , b ) 45 , c ) 55 , d ) 70 , e ) 30 | b | subtract(175, add(add(20, 80), divide(add(20, subtract(210, add(add(80, 70), 20))), 2))) | add(n7,n13)|add(n7,n10)|add(n13,#1)|subtract(n3,#2)|add(n13,#3)|divide(#4,n4)|add(#0,#5)|subtract(n1,#6) | physics |
rs . 1500 is divided into two parts such that if one part is invested at 6 % and the other at 5 % the whole annual interest from both the sum is rs . 75 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 1500 - x ) * 6 * 1 ] / 100 = 75 5 x / 100 + 90 – 6 x / 100 = 75 x / 100 = 15 = > x = 1500 . answer : b" | a ) 388 , b ) 1500 , c ) 277 , d ) 500 , e ) 271 | b | multiply(add(5, 6), const_100) | add(n1,n2)|multiply(#0,const_100)| | gain |
a sum of money at simple interest amounts to $ 980 in 3 years and to $ 1024 in 4 years . the sum is : | "b $ 848 s . i . for 1 year = $ ( 1024 - 980 ) = $ 44 . s . i . for 3 years = $ ( 44 x 3 ) = $ 132 . principal = $ ( 980 - 132 ) = $ 848 ." | a ) $ 153 , b ) $ 848 , c ) $ 398 , d ) $ 549 , e ) $ 675 | b | subtract(980, divide(multiply(subtract(1024, 980), 3), 4)) | subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)| | gain |
the perimeter of a rectangular yard is completely surrounded by a fence that measures 18 meters . what is the length of the yard if the area of the yard is 20 meters squared ? | "perimeter of rectangular yard = 2 ( l + b ) = 18 - - > l + b = 9 area = l * b = 20 b = 9 - l l ( 9 - l ) = 20 29 l - l ^ 2 = 20 l ^ 2 - 9 l + 20 = 0 upon simplifying we get l = 5 or 4 . only 5 is there in the answer choice . answer : d" | a ) 8 , b ) 2 , c ) 7 , d ) 5 , e ) 1 | d | subtract(const_4, const_3) | subtract(const_4,const_3)| | geometry |
the shopkeeper increased the price of a product by 25 % so that customer finds it difficult to purchase the required amount . but somehow the customer managed to purchase only 68 % of the required amount . what is the net difference in the expenditure on that product ? | "quantity x rate = price 1 x 1 = 1 0.68 x 1.25 = 0.85 decrease in price = ( 0.15 / 1 ) × 100 = 15 % d )" | a ) 12.5 % , b ) 13 % , c ) 16 % , d ) 15 % , e ) 19 % | d | divide(multiply(subtract(multiply(const_100, const_100), multiply(add(const_100, 25), 68)), const_100), multiply(const_100, const_100)) | add(n0,const_100)|multiply(const_100,const_100)|multiply(n1,#0)|subtract(#1,#2)|multiply(#3,const_100)|divide(#4,#1)| | general |
a store reduced the price of all items in the store by 10 % on the first day and by another 14 % on the second day . the price of items on the second day was what percent of the price before the first reduction took place ? | "consider price of the all items as $ 100 after a initial reduction of 10 % price becomes = 0.9 * 100 = $ 90 after the final reduction of 14 % price becomes = 0.86 * 90 = $ 77.4 price of all items on second day is 77.4 % of price on first day correct answer option c" | a ) 80.0 , b ) 80.9 , c ) 77.4 , d ) 81.1 , e ) 81.9 | c | multiply(multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 14), const_100)), const_100) | subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|multiply(#4,const_100)| | gain |
if 25 % of ( x - y ) = 15 % of ( x + y ) , then what percent of x is y ? | "explanation : solution : 25 % of ( x - y ) = 15 % of ( x + y ) 25 ( x - y ) / 100 = 15 ( x + y ) / 100 5 ( x - y ) = 3 ( x + y ) x = 4 y . . ' . required percentage = ( y * 100 / x ) % = y * 100 / 4 y = 25 % answer : d" | a ) 20 % , b ) 15 % , c ) 30 % , d ) 25 % , e ) none of these | d | multiply(divide(subtract(25, 15), add(25, 15)), const_100) | add(n0,n1)|subtract(n0,n1)|divide(#1,#0)|multiply(#2,const_100)| | general |
the average weight of 3 men a , b and c is 84 kg . the average weight becomes 80 kg when d joins them . if e whose weight is 3 kg more than d joins the group replacing a , then the average weight of b , c , d and e becomes 79 kg . the weight of a is : | wt of abc = 84 * 3 = 252 kg wt of abcd = 80 * 4 = 320 kg wt of d = 68 kg wt of e = 71 kg wt of abcde = 320 + 71 = 391 kg wt of bcde = 79 * 4 = 316 kg wt of a = 391 - 316 = 75 kg answer : c | a ) 65 kg , b ) 70 kg , c ) 75 kg , d ) 80 kg , e ) 85 kg | c | subtract(multiply(80, const_4), subtract(multiply(79, const_4), add(subtract(multiply(80, const_4), multiply(84, 3)), 3))) | multiply(n2,const_4)|multiply(n4,const_4)|multiply(n0,n1)|subtract(#0,#2)|add(n0,#3)|subtract(#1,#4)|subtract(#0,#5) | general |
the ratio of the present ages of two friends is 2 : 3 and 6 years back , the ratio was 1 : 3 . what will be the ratio of their ages after 4 years ? | let the ages be 2 x , 3 x 6 yrs back , so 2 x - 6 / 3 x - 6 = 1 / 3 x = 4 after 4 yrs 2 x + 43 x + 4 2 ( 4 ) + 43 ( 4 ) + 4 12 : 16 3 : 4 answer : b | a ) 1 : 4 , b ) 3 : 4 , c ) 1 : 2 , d ) 3 : 1 , e ) 3 : 2 | b | divide(add(multiply(divide(2, 3), divide(subtract(multiply(6, 3), 6), subtract(multiply(multiply(const_1, const_3), divide(2, 3)), 1))), 4), add(divide(subtract(multiply(6, 3), 6), subtract(multiply(multiply(const_1, const_3), divide(2, 3)), 1)), 4)) | divide(n0,n1)|multiply(n1,n2)|multiply(const_1,const_3)|multiply(#0,#2)|subtract(#1,n2)|subtract(#3,n3)|divide(#4,#5)|add(n5,#6)|multiply(#0,#6)|add(n5,#8)|divide(#9,#7) | other |
the perimeter of a triangle is 24 cm and the inradius of the triangle is 2.5 cm . what is the area of the triangle ? | "area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 2.5 * 24 / 2 = 30 cm 2 answer : a" | a ) 30 cm 2 , b ) 85 cm 2 , c ) 65 cm 2 , d ) 45 cm 2 , e ) 35 cm 2 | a | triangle_area(2.5, 24) | triangle_area(n0,n1)| | geometry |
if 85 % of the population of an ant colony is red , and of these 45 % are females , then what % of the total ant population are male red ants ? | 55 % are males percentage of male red ants is 55 * . 85 = 46.75 answer : a | a ) 46.75 , b ) 40 , c ) 33.66 , d ) 66.66 , e ) 66.86 | a | multiply(multiply(subtract(const_1, divide(45, const_100)), divide(85, const_100)), const_100) | divide(n0,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(#0,#2)|multiply(#3,const_100) | gain |
if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 54 / r years . if pat ' s parents invested $ 6,000 in a long - term bond that pays 6 percent interest , compounded annually , what will be the approximate total amount of the investment 18 years ... | "since investment doubles in 54 / r years , then for r = 6 it ' ll double in 54 / 6 = ~ 9 years ( we are not asked about the exact amount so such an approximation will do ) . thus after 18 years investment will become $ 6,000 * 2 = $ 12,000 . answer : c ." | a ) $ 20000 , b ) $ 15000 , c ) $ 12000 , d ) $ 10000 , e ) $ 9000 | c | divide(multiply(multiply(add(const_2, const_3), const_1000), 6), const_2) | add(const_2,const_3)|multiply(#0,const_1000)|multiply(n2,#1)|divide(#2,const_2)| | general |
a train 210 metres long is moving at a speed of 25 kmph . it will cross a man coming from the opposite direction at a speed of 2 km per hour in : | "relative speed = ( 25 + 2 ) km / hr = 27 km / hr = ( 27 × 5 / 18 ) m / sec = 15 / 2 m / sec . time taken by the train to pass the man = ( 210 × 2 / 15 ) sec = 28 sec answer : a" | a ) 28 sec , b ) 32 sec , c ) 36 sec , d ) 38 sec , e ) 40 sec | a | multiply(const_3600, divide(divide(210, const_1000), add(25, 2))) | add(n1,n2)|divide(n0,const_1000)|divide(#1,#0)|multiply(#2,const_3600)| | physics |
in 3 annual examinations , of which the aggregate marks of each was 500 , a student secured average marks 45 % and 55 % in the first and the second yearly examinations respectively . to secure 40 % average total marks , it is necessary for him in third yearly examination to secure marks : | total marks : 1500 for three exams 40 % of 1500 = 600 first exam marks = 45 % of 500 = 225 second exam marks = 55 % of 500 = 275 let x be the third exam marks 225 + 275 + x = 600 x = 100 answer : a | a ) 100 , b ) 350 , c ) 400 , d ) 450 , e ) 500 | a | subtract(divide(multiply(multiply(3, 500), 40), const_100), add(divide(multiply(500, 55), const_100), divide(multiply(500, 45), const_100))) | multiply(n0,n1)|multiply(n1,n3)|multiply(n1,n2)|divide(#1,const_100)|divide(#2,const_100)|multiply(n4,#0)|add(#3,#4)|divide(#5,const_100)|subtract(#7,#6) | general |
a man has some hens and cows . if the number of heads be 46 and the number of feet equals 140 , then the number of hens will be : | "let hens be x and cows be y now , feet : x * 2 + y * 4 = 140 heads : x * 1 + y * 1 = 46 implies , 2 x + 4 y = 140 and x + y = 46 solving these two equations , we get x = 22 and y = 24 therefore , hens are 22 . answer : a" | a ) 22 , b ) 23 , c ) 24 , d ) 26 , e ) 28 | a | divide(subtract(multiply(46, const_4), 140), const_2) | multiply(n0,const_4)|subtract(#0,n1)|divide(#1,const_2)| | general |
two negative numbers are multiplied to give a product of 48 . if the lesser number is 10 less than thrice the greater number , what is the greater number ? | test the options . the options give you the greater number . ( a ) - 6 triple of - 6 is - 18 and 10 less is - 8 . - 8 * - 6 = 48 ( correct ) correct answer ( a ) | a ) - 6 , b ) - 8 , c ) - 5 , d ) - 9 , e ) - 10 | a | divide(subtract(negate(sqrt(subtract(power(10, const_2), multiply(multiply(negate(48), const_3), const_4)))), 10), multiply(const_2, const_3)) | multiply(const_2,const_3)|negate(n0)|power(n1,const_2)|multiply(#1,const_3)|multiply(#3,const_4)|subtract(#2,#4)|sqrt(#5)|negate(#6)|subtract(#7,n1)|divide(#8,#0) | general |
yesterday ' s closing prices of 2,860 different stocks listed on a certain stock exchange were all different from today ' s closing prices . the number of stocks that closed at a higher price today than yesterday was 20 percent greater than the number that closed at a lower price . how many of the stocks closed at a hi... | "lets consider the below - the number of stocks that closed at a higher price = h the number of stocks that closed at a lower price = l we understand from first statement - > h + l = 2860 - - - - ( 1 ) we understand from second statement - > h = ( 120 / 100 ) l = > h = 1.2 l - - - - ( 2 ) solve eq ( 1 ) ( 2 ) to get h ... | a ) 484 , b ) 726 , c ) 1,100 , d ) 1,320 , e ) 1,560 | e | multiply(divide(subtract(subtract(multiply(20, const_100), const_10), const_10), add(add(const_1, divide(20, const_100)), const_1)), add(const_1, divide(20, const_100))) | divide(n1,const_100)|multiply(n1,const_100)|add(#0,const_1)|subtract(#1,const_10)|add(#2,const_1)|subtract(#3,const_10)|divide(#5,#4)|multiply(#2,#6)| | gain |
from a square piece of a paper having each side equal to 10 cm , the largest possible circle is being cut out . the ratio of the area of the circle to the area of the original square is nearly : | area of the square = ( 10 ) 2 = 100 cm 2 area of the circle = 22 ⁄ 7 × ( 5 ) 2 = 22 × 25 / 7 required ratio = 22 × 25 / 7 × 100 = 22 / 28 = 11 / 14 = 0.785 ≈ 0.8 = 4 ⁄ 5 answer a | ['a ) 4 ⁄ 5', 'b ) 3 ⁄ 5', 'c ) 5 ⁄ 6', 'd ) 6 ⁄ 7', 'e ) none of these'] | a | multiply(divide(circle_area(divide(10, const_2)), square_area(10)), const_100) | divide(n0,const_2)|square_area(n0)|circle_area(#0)|divide(#2,#1)|multiply(#3,const_100) | geometry |
light glows for every 13 seconds . how many times did it between 1 : 57 : 58 and 3 : 20 : 47 am | the diff in sec between 1 : 57 : 58 and 3 : 20 : 47 is 4969 sec , 4969 / 13 = 382 . so total 383 times light ll glow answer : c | a ) 381 , b ) 382 , c ) 383 , d ) 384 , e ) 385 | c | divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 13) | add(n6,const_2)|add(const_2,const_60)|add(n5,#1)|multiply(#2,const_60)|add(#0,#3)|divide(#4,n0) | physics |
if 7 men and 2 boys working together , can do 6 times as much work per hour as a man and a boy together . find the ratio of the work done by a man and that of a boy for a given time ? | 7 m + 2 b = 6 ( 1 m + 1 b ) 7 m + 2 b = 6 m + 6 b 1 m = 4 b the required ratio of work done by a man and a boy = 4 : 1 answer : d | a ) 2 : 5 , b ) 2 : 3 , c ) 1 : 8 , d ) 4 : 1 , e ) 3 : 2 | d | divide(add(2, 2), const_1) | add(n1,n1)|divide(#0,const_1) | physics |
if x is 20 percent greater than 52 , then x = | "x is 20 % greater than 52 means x is 1.2 times 52 ( in other words 52 + 20 / 100 * 52 = 1.2 * 52 ) therefore , x = 1.2 * 88 = 62.4 answer : b" | a ) 68 , b ) 62.4 , c ) 86 , d ) 72.8 , e ) 108 | b | add(52, multiply(divide(20, const_100), 52)) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)| | general |
two stations p and q are 245 km apart on a straight track . one train starts from p at 7 a . m . and travels towards q at 20 kmph . another train starts from q at 8 a . m . and travels towards p at a speed of 25 kmph . at what time will they meet ? | "assume both trains meet after x hours after 7 am distance covered by train starting from p in x hours = 20 x km distance covered by train starting from q in ( x - 1 ) hours = 25 ( x - 1 ) total distance = 245 = > 20 x + 25 ( x - 1 ) = 245 = > 45 x = 270 = > x = 6 means , they meet after 6 hours after 7 am , ie , they ... | a ) 10 am , b ) 12 am , c ) 10.30 am , d ) 12.30 am , e ) 1 pm | e | add(divide(add(245, 25), add(20, 25)), 7) | add(n0,n4)|add(n2,n4)|divide(#0,#1)|add(n1,#2)| | physics |
a girl was asked to multiply a certain number by 43 . she multiplied it by 34 and got his answer less than the correct one by 1215 . find the number to be multiplied . | let the required number be x . then , 43 x – 34 x = 1215 or 9 x = 1215 or x = 135 . required number = 135 answer : c | a ) 130 , b ) 132 , c ) 135 , d ) 136 , e ) 138 | c | divide(1215, subtract(43, 34)) | subtract(n0,n1)|divide(n2,#0) | general |
there is 60 lit of milk and water in which milk forms 84 % . howmuch water must be added to this solution to make it solution in which milk forms 64 % | "60 * 84 / 100 = 50.40 lit milk that is 9.60 lit water let x lit water will be added then ( 60 + x ) * 64 / 100 = 50.40 so x = 18.75 answer : a" | a ) 18.75 , b ) 19.75 , c ) 20.75 , d ) 21.75 , e ) 22.75 | a | subtract(multiply(divide(const_100, 64), divide(multiply(60, 84), const_100)), 60) | divide(const_100,n2)|multiply(n0,n1)|divide(#1,const_100)|multiply(#0,#2)|subtract(#3,n0)| | gain |
if x + y = - 10 , and x = 25 / y , what is the value of x ^ 2 + y ^ 2 ? | "x ^ 2 + y ^ 2 should make you think of these formulas : ( x + y ) ( x + y ) = x ^ 2 + y ^ 2 + 2 xy we already know ( x + y ) = - 10 and x * y = 25 ( x + y ) ( x + y ) = ( - 10 ) ( - 10 ) = x ^ 2 + y ^ 2 + 2 * ( 25 ) x ^ 2 + y ^ 2 = 100 - 50 = 50 answer : c" | a ) 55 , b ) 65 , c ) 50 , d ) 75 , e ) 85 | c | subtract(power(10, 2), multiply(2, 25)) | multiply(n1,n2)|power(n0,n2)|subtract(#1,#0)| | general |
a type r machine can complete a job in 5 hours and a type b machine can complete the job in 7 hours . how many hours will it take 2 type r machines and 3 type b machines working together and independently to complete the job ? | now d should be the answer . r need 5 hours to complete and b needs 7 hours to compete so 2 r + 3 b will complete 2 / 5 + 3 / 7 or 29 / 35 portion of the job in 1 hour so the whole job will take 35 / 29 hours . . . . = d | a ) 1 / 5 , b ) 29 / 35 , c ) 5 / 6 , d ) 35 / 29 , e ) 35 / 12 | d | divide(const_1, add(divide(2, 5), divide(3, 7))) | divide(n2,n0)|divide(n3,n1)|add(#0,#1)|divide(const_1,#2) | physics |
the speed of a boat in still water in 42 km / hr and the rate of current is 6 km / hr . the distance travelled downstream in 44 minutes is : | "speed downstream = ( 42 + 6 ) = 48 kmph time = 44 minutes = 44 / 60 hour = 11 / 15 hour distance travelled = time × speed = 11 / 15 × 48 = 35.2 km answer : c" | a ) 86.6 km , b ) 46.6 km , c ) 35.2 km , d ) 35.6 km , e ) 26.6 km | c | multiply(add(42, 6), divide(44, const_60)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics |
a basketball team composed of 12 players scored 100 points in a particular contest . if none of the individual players scored fewer than 7 points , what is the greatest number of points l that an individual player might have scored ? | "general rule for such kind of problems : to maximize one quantity , minimize the others ; to minimize one quantity , maximize the others . thus to maximize the number of points of one particular player minimize the number of points of all other 11 players . minimum number of points for a player is 7 , so the minimum n... | a ) 7 , b ) 13 , c ) 16 , d ) 21 , e ) 23 | e | add(subtract(100, multiply(12, 7)), 7) | multiply(n0,n2)|subtract(n1,#0)|add(n2,#1)| | general |
find the value of log y ( x 4 ) if logx ( y 3 ) = 2 | logx ( y 3 ) = 2 : given x 2 = y 3 : rewrite in exponential form x 4 = y 6 : square both sides x 4 = y 6 : rewrite the above using the log base y logy ( x 4 ) = logy ( y 6 ) = 6 correct answer c | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 9 | c | add(4, 2) | add(n0,n2) | general |
arun and tarun can do a work in 10 days . after 4 days tarun went to his village . how many days are required to complete the remaining work by arun alone . arun can do the work alone in 70 days . | "they together completed 4 / 10 work in 4 days . balance 6 / 10 work will be completed by arun alone in 70 * 6 / 10 = 42 days . answer : e" | a ) 16 days . , b ) 17 days . , c ) 18 days . , d ) 19 days . , e ) 42 days . | e | subtract(70, multiply(divide(70, 10), 4)) | divide(n2,n0)|multiply(n1,#0)|subtract(n2,#1)| | physics |
a 100 - litre mixture of milk and water contains 36 litres of milk . ' x ' litres of this mixture is removed and replaced with an equal quantum of water . if the process is repeated once , then the concentration of the milk stands reduced at 16 % . what is the value of x ? | "working formula . . . initial concentration * initial volume = final concentration * final volume . let x is the part removed from 100 lts . 36 % ( 1 - x / 100 ) ^ 2 = 16 % * 100 % ( 1 - x / 100 ) ^ 2 = 16 / 36 - - - - - - > ( 1 - x / 100 ) ^ 2 = ( 4 / 6 ) ^ 2 100 - x = 400 / 6 x = 33.33 . . . ans e" | a ) 37.5 litres , b ) 36.67 litres , c ) 37.67 litres , d ) 36.5 litres , e ) 33.33 litres | e | multiply(100, subtract(const_1, sqrt(divide(16, 36)))) | divide(n2,n1)|sqrt(#0)|subtract(const_1,#1)|multiply(n0,#2)| | general |
a man â € ™ s current age is ( 2 / 5 ) of the age of his father . after 5 years , he will be ( 1 / 2 ) of the age of his father . what is the age of father at now ? | "let , father â € ™ s current age is a years . then , man â € ™ s current age = [ ( 2 / 5 ) a ] years . therefore , [ ( 2 / 5 ) a + 5 ] = ( 1 / 2 ) ( a + 5 ) 2 ( 2 a + 25 ) = 5 ( a + 8 ) a = 25 b" | a ) 40 , b ) 25 , c ) 38 , d ) 50 , e ) 39 | b | divide(subtract(5, multiply(5, divide(1, 2))), subtract(divide(1, 2), divide(2, 5))) | divide(n3,n0)|divide(n0,n1)|multiply(n2,#0)|subtract(#0,#1)|subtract(n2,#2)|divide(#4,#3)| | general |
after an ice began to melt out from the freezer , in the first hour lost 3 / 4 , in the second hour lost 3 / 4 of its remaining . if after two hours , the volume is 0.75 cubic inches , what is the original volume of the cubic ice , in cubic inches ? | "let initial volume of ice be = x ice remaining after 1 hour = x - 0.75 x = 0.25 x ice remaining after 2 hour = ( 1 / 4 ) x - ( 3 / 4 * 1 / 4 * x ) = ( 1 / 16 ) x ( 1 / 16 ) x = 0.75 x = 12 alternate solution : try to backsolve . initial volume = 12 after one hour - - > ( 1 / 4 ) 12 = 3 after two hours - - > ( 1 / 4 ) ... | a ) 2.5 , b ) 3.0 , c ) 4.0 , d ) 6.5 , e ) 12.0 | e | divide(divide(0.75, const_0_25), const_0_25) | divide(n4,const_0_25)|divide(#0,const_0_25)| | physics |
the average height of 20 girls out of a class of 50 is 142 cm . and that of the remaining girls is 149 cm . the average height of the whole class is : | "explanation : average height of the whole class = ( 20 × 142 + 30 × 149 / 50 ) = 146.2 cms answer a" | a ) 146.2 cms , b ) 146.5 cms , c ) 146.9 cms , d ) 142.2 cms , e ) 136.2 cms | a | divide(add(multiply(142, 20), multiply(149, const_10)), 50) | multiply(n0,n2)|multiply(n3,const_10)|add(#0,#1)|divide(#2,n1)| | general |
a shopkeeper bought 600 oranges and 400 bananas . he found 15 % of oranges and 3 % of bananas were rotten . find the percentage of fruits in good condition ? | total number of fruits shopkeeper bought = 600 + 400 = 1000 number of rotten oranges = 15 % of 600 = 15 / 100 × 600 = 9000 / 100 = 90 number of rotten bananas = 3 % of 400 = 12 therefore , total number of rotten fruits = 90 + 12 = 102 therefore number of fruits in good condition = 1000 - 102 = 898 therefore percentage ... | a ) 92.5 % , b ) 89.8 % , c ) 85.2 % , d ) 96.8 % , e ) 78.9 % | b | multiply(divide(subtract(add(600, 400), add(multiply(600, divide(15, const_100)), multiply(400, divide(3, const_100)))), add(600, 400)), const_100) | add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n1,#2)|add(#3,#4)|subtract(#0,#5)|divide(#6,#0)|multiply(#7,const_100) | gain |
if taxi fares were $ 1.00 for the first 1 / 5 mile and $ 0.40 for each 1 / 5 mile there after , then the taxi fare for a 3 - mile ride was | "in 3 miles , initial 1 / 5 mile charge is $ 1 rest of the distance = 3 - ( 1 / 5 ) = 14 / 5 rest of the distance charge = 14 ( 0.4 ) = $ 5.6 ( as the charge is 0.4 for every 1 / 5 mile ) = > total charge for 3 miles = 1 + 5.6 = 6.6 answer is e ." | a ) $ 1.56 , b ) $ 2.40 , c ) $ 3.80 , d ) $ 4.20 , e ) $ 6.60 | e | add(1.00, multiply(subtract(divide(1.00, divide(1, 5)), 1), 0.40)) | divide(n1,n2)|divide(n6,#0)|subtract(#1,n1)|multiply(n3,#2)|add(n0,#3)| | general |
bruce purchased 9 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | "cost of 9 kg grapes = 70 × 9 = 630 . cost of 9 kg of mangoes = 55 × 9 = 495 total cost he has to pay = 630 + 495 = 1125 e" | a ) a ) 1040 , b ) b ) 1050 , c ) c ) 1055 , d ) d ) 1065 , e ) e ) 1125 | e | add(multiply(9, 70), multiply(9, 55)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | gain |
sum of two numbers is 35 . two times of the first exceeds by 5 from the three times of the other . then the numbers will be ? | "explanation : x + y = 35 2 x – 3 y = 5 x = 22 y = 13 d )" | a ) a ) 5 , b ) b ) 9 , c ) c ) 11 , d ) d ) 22 , e ) e ) 15 | d | subtract(35, divide(subtract(35, divide(5, const_2)), const_2)) | divide(n1,const_2)|subtract(n0,#0)|divide(#1,const_2)|subtract(n0,#2)| | general |
the batting average of a particular batsman is 60 runs in 46 innings . if the difference in his highest and lowest score is 180 runs and his average excluding these two innings is 58 runs , find his highest score . | "explanation : total runs scored by the batsman = 60 * 46 = 2760 runs now excluding the two innings the runs scored = 58 * 44 = 2552 runs hence the runs scored in the two innings = 2760 â € “ 2552 = 208 runs . let the highest score be x , hence the lowest score = x â € “ 180 x + ( x - 180 ) = 208 2 x = 388 x = 194 runs... | a ) 179 , b ) 194 , c ) 269 , d ) 177 , e ) 191 | b | divide(add(180, subtract(multiply(60, 46), multiply(58, subtract(46, const_2)))), const_2) | multiply(n0,n1)|subtract(n1,const_2)|multiply(n3,#1)|subtract(#0,#2)|add(n2,#3)|divide(#4,const_2)| | general |
the average weight of 19 students is 15 kg . by the admission of a new student the average weight is reduced to 14.8 kg . the weight of the new student is ? | answer weight of new student = total weight of all 20 students - total weight of initial 19 students = ( 20 x 14.8 - 19 x 15 ) kg = 11 kg . correct option : c | a ) 10.6 kg , b ) 10.8 kg , c ) 11 kg , d ) 14.9 kg , e ) none | c | subtract(multiply(add(19, const_1), 14.8), multiply(19, 15)) | add(n0,const_1)|multiply(n0,n1)|multiply(n2,#0)|subtract(#2,#1) | general |
john has $ 1,600 at the beginning of his trip , after spending money , he still has exactly $ 800 less than he spent on the trip . how much money does john still have ? | "let the money spent be x money he is left with after spending = x - 800 total money - - > x + ( x - 800 ) = 1600 solving for x will give x = 1200 , therefore the money he is left with = x - 800 = 1200 - 800 = 400 answer : b" | a ) $ 200 , b ) $ 400 , c ) $ 600 , d ) $ 800 , e ) $ 1,200 | b | subtract(add(multiply(const_100, const_10), 800), divide(add(800, add(multiply(const_100, const_10), 800)), const_2)) | multiply(const_10,const_100)|add(n1,#0)|add(n1,#1)|divide(#2,const_2)|subtract(#1,#3)| | general |
how many 3 digit number contain number 4 ? | "total 3 digit no . = 9 * 10 * 10 = 900 not containing 4 = 8 * 9 * 9 = 648 total 3 digit number contain 4 = 900 - 648 = 252 answer : d" | a ) 352 , b ) 268 , c ) 236 , d ) 252 , e ) 354 | d | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(3, 3)), multiply(const_4, const_2))), const_10) | multiply(n0,n0)|multiply(const_2,const_4)|subtract(const_1000,const_10)|multiply(#0,const_10)|multiply(#3,#1)|subtract(#2,#4)|add(#5,const_10)| | general |
the total age of a and b is 16 years more than the total age of b and c . c is how many year younger than a | "explanation : given that a + b = 16 + b + c = > a ? c = 16 + b ? b = 16 = > c is younger than a by 16 years answer : option b" | a ) 11 , b ) 16 , c ) 13 , d ) 14 , e ) 15 | b | multiply(16, const_1) | multiply(n0,const_1)| | general |
how many positive integers less than 14 can be expressed as the sum of a positive multiple of 2 and a positive multiple of 3 ? | "the number = 2 a + 3 b < 20 when a = 1 , b = 1 , 2 , 3 , 4 , 5 - > 2 a = 2 ; 3 b = 3 , 6 , 9 - > the number = 5 , 8 , 11 - - > 3 numbers when a = 2 , b = 1 , 2,3 - > . . . . - - > 3 numbers when a = 3 , b = 1,2 , 3,4 - - > . . . . - - > 2 numbers total number is already 8 . look at the answer there is no number greate... | a ) 14 , b ) 13 , c ) 12 , d ) 11 , e ) 8 | e | subtract(subtract(subtract(14, 2), const_4), const_1) | subtract(n0,n1)|subtract(#0,const_4)|subtract(#1,const_1)| | general |
if 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days , the time taken by 15 men and 20 boys in doing the same type of work will be : | let 1 man ' s 1 day ' s work = x and 1 boy ' s 1 day ' s work = y . then , 6 x + 8 y = 1 / 10 and 26 x + 48 y = 1 / 2 solving these two equations , we get : x = 1 / 100 and y = 1 / 200 ( 15 men + 20 boy ) ' s 1 day ' s work = ( 15 / 100 + 20 / 200 ) = 1 / 4 . 15 men and 20 boys can do the work in 4 days . answer : a | a ) 4 days , b ) 5 days , c ) 6 days , d ) 8 days , e ) 2 days | a | divide(multiply(add(6, const_4), 10), add(15, 10)) | add(n0,const_4)|add(n2,n6)|multiply(n2,#0)|divide(#2,#1) | physics |
a batsman makes a score of 87 runs in the 17 th match and thus increases his average by 3 . find his average after 17 th match | "explanation : let the average after 17 th match is x then the average before 17 th match is x - 3 so 16 ( x - 3 ) + 87 = 17 x = > x = 87 - 48 = 39 option d" | a ) 36 , b ) 37 , c ) 38 , d ) 39 , e ) 35 | d | add(subtract(87, multiply(17, 3)), 3) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general |
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 $ 1.20 at the candy store . what is this student ’ s weekly allowance ? | let x be the value of the weekly allowance . ( 2 / 3 ) ( 2 / 5 ) x = 120 cents ( 4 / 15 ) x = 120 x = $ 4.50 the answer is d . | a ) $ 3.50 , b ) $ 4.00 , c ) $ 4.25 , d ) $ 4.50 , e ) $ 5.00 | d | divide(multiply(multiply(3, 5), 1.2), const_4) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_4) | general |
a man rows his boat 78 km downstream and 50 km upstream , taking 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 78 / ( 2 ) = 39 kmph speed upstream = d / t = 50 / ( 2 ) = 25 kmph the speed of the stream = ( 39 - 25 ) / 2 = 7 kmph answer : d" | a ) 76 kmph , b ) 6 kmph , c ) 14 kmph , d ) 7 kmph , e ) 4 kmph | d | divide(subtract(divide(78, 2), divide(50, 2)), const_2) | divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)| | physics |
the difference of two numbers is 1465 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder . what is the smaller number ? | "let the smaller number be x . then larger number = ( x + 1465 ) . x + 1465 = 6 x + 15 5 x = 1450 x = 290 smaller number = 290 . c )" | a ) a ) 270 , b ) b ) 280 , c ) c ) 290 , d ) d ) 300 , e ) e ) 310 | c | divide(add(1465, 15), subtract(6, const_1)) | add(n0,n2)|subtract(n1,const_1)|divide(#0,#1)| | general |
if it takes a machine 3 ⁄ 5 minute to produce one item , how many items will it produce in 2 hours ? | "1 item takes 3 / 5 min so it takes 120 min to produce x 3 x / 5 = 120 the x = 200 answer : d" | a ) 1 ⁄ 3 , b ) 4 ⁄ 3 , c ) 80 , d ) 200 , e ) 180 | d | divide(multiply(2, const_60), divide(3, 5)) | divide(n0,n1)|multiply(n2,const_60)|divide(#1,#0)| | physics |
$ 364 is divided among a , b , and c so that a receives half as much as b , and b receives half as much as c . how much money is c ' s share ? | "let the shares for a , b , and c be x , 2 x , and 4 x respectively . 7 x = 364 x = 52 4 x = 208 the answer is c ." | a ) $ 200 , b ) $ 204 , c ) $ 208 , d ) $ 212 , e ) $ 216 | c | multiply(divide(364, add(add(divide(const_1, const_2), const_1), const_2)), const_2) | divide(const_1,const_2)|add(#0,const_1)|add(#1,const_2)|divide(n0,#2)|multiply(#3,const_2)| | general |
the price of a jacket is reduced by 25 % . during a special sale the price of the jacket is reduced another 20 % . by approximately what percent must the price of the jacket now be increased in order to restore it to its original amount ? | "1 ) let the price of jacket initially be $ 100 . 2 ) then it is decreased by 25 % , therefore bringing down the price to $ 75 . 3 ) again it is further discounted by 20 % , therefore bringing down the price to $ 60 . 4 ) now 60 has to be added byx % in order to equal the original price . 60 + ( x % ) 60 = 100 . solvin... | a ) 32.5 , b ) 35 , c ) 48 , d ) 65 , e ) 66.67 | e | multiply(const_100, divide(subtract(const_100, subtract(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(20, const_100)))), subtract(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(20, const_100))))) | divide(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|subtract(#1,#2)|subtract(const_100,#3)|divide(#4,#3)|multiply(#5,const_100)| | gain |
two consultants can type up a report in 12.5 hours and edit it in 7.5 hours . if mary needs 30 hours to type the report and jim needs 12 hours to edit it alone , how many e hours will it take if jim types the report and mary edits it immediately after he is done ? | "break down the problem into two pieces : typing and editing . mary needs 30 hours to type the report - - > mary ' s typing rate = 1 / 30 ( rate reciprocal of time ) ( point 1 in theory below ) ; mary and jim can type up a report in 12.5 and - - > 1 / 30 + 1 / x = 1 / 12.5 = 2 / 25 ( where x is the time needed for jim ... | a ) 41.4 , b ) 34.1 , c ) 13.4 , d ) 12.4 , e ) 10.8 | a | add(inverse(subtract(divide(const_1, 12.5), divide(const_1, 30))), inverse(subtract(divide(const_1, 7.5), divide(const_1, 12)))) | divide(const_1,n0)|divide(const_1,n2)|divide(const_1,n1)|divide(const_1,n3)|subtract(#0,#1)|subtract(#2,#3)|inverse(#4)|inverse(#5)|add(#6,#7)| | physics |
a man saves 20 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 10 % , he is only able to save rs . 500 per month . what is his monthly salary ? | income = rs . 100 expenditure = rs . 80 savings = rs . 20 present expenditure 80 + 80 * ( 10 / 100 ) = rs . 88 present savings = 100 – 88 = rs . 12 if savings is rs . 12 , salary = rs . 100 if savings is rs . 500 , salary = 100 / 12 * 500 = 4167 answer : c | a ) rs . 4500 , b ) rs . 4000 , c ) rs . 4167 , d ) rs . 4200 , e ) rs . 3000 | c | divide(multiply(500, const_100), subtract(const_100, add(subtract(const_100, 20), multiply(subtract(const_100, 20), divide(10, const_100))))) | divide(n1,const_100)|multiply(n2,const_100)|subtract(const_100,n0)|multiply(#0,#2)|add(#3,#2)|subtract(const_100,#4)|divide(#1,#5) | general |
a started a business with an investment of rs . 70000 and after 6 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 78000 , then the share of a is ? | "ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 6 ) = 7 : 6 total profit = rs . 78000 share of b = 7 / 13 ( 78000 ) = rs . 42000 answer : b" | a ) s . 42028 , b ) s . 42000 , c ) s . 42003 , d ) s . 42029 , e ) s . 24029 | b | subtract(78000, multiply(const_60, const_100)) | multiply(const_100,const_60)|subtract(n3,#0)| | gain |
what is the minimum value of | x - 4 | + | x + 7 | + | x - 5 | ? | "a can not be the answer as all the three terms are in modulus and hence the answer will be non negative . | x - 4 | > = 0 - - > minimum occurs at x = 4 | x + 7 | > = 0 - - > minimum occurs at x = - 7 | x - 5 | > = 0 - - > minimum occurs at x = 5 x = - 7 - - > result = 11 + 0 + 12 = 23 . also any negative value will pu... | a ) - 3 , b ) 3 , c ) 5 , d ) - 12 , e ) 12 | e | add(7, 5) | add(n1,n2)| | general |
the average of 1 st 3 of 4 numbers is 16 and of the last 3 are 15 . if the sum of the first and the last number is 13 . what is the last numbers ? | "a + b + c = 48 b + c + d = 45 a + d = 13 a – d = 3 a + d = 13 2 d = 10 d = 5" | a ) 2 , b ) 4 , c ) 6 , d ) 5 , e ) 7 | d | subtract(subtract(multiply(3, 16), add(subtract(13, 16), 3)), 16) | multiply(n1,n3)|subtract(n6,n3)|add(n1,#1)|subtract(#0,#2)|subtract(#3,n3)| | general |
mahesh marks an article 15 % above the cost price of rs . 540 . what must be his discount percentage if he sells it at rs . 456 ? | "cp = rs . 540 , mp = 540 + 15 % of 540 = rs . 621 sp = rs . 456 , discount = 621 - 456 = 165 discount % = 165 / 621 * 100 = 26.57 % answer : b" | a ) 18 % , b ) 26.57 % , c ) 20 % , d ) 19 % , e ) none of these | b | divide(multiply(subtract(multiply(540, divide(add(const_100, 15), const_100)), 456), const_100), multiply(540, divide(add(const_100, 15), const_100))) | add(n0,const_100)|divide(#0,const_100)|multiply(n1,#1)|subtract(#2,n2)|multiply(#3,const_100)|divide(#4,#2)| | gain |
the h . c . f . of two numbers is 11 and their l . c . m . is 693 . if one of the numbers is 77 , find the other . | "other number = 11 x 693 / 77 = 99 answer is a ." | a ) 99 , b ) 97 , c ) 95 , d ) 91 , e ) 96 | a | multiply(11, 77) | multiply(n0,n2)| | physics |
kate and danny each have $ 10 . together , they flip a fair coin 5 times . every time the coin lands on heads , kate gives danny $ 1 . every time the coin lands on tails , danny gives kate $ 1 . after the 5 coin flips , what is the probability that kate has more than $ 10 but less than $ 15 ? | the probability of the coin landing tails up either 3 or 4 times = p ( 3 t ) + p ( 4 t ) binomial distribution formula : nck p ^ k ( 1 - p ) ^ ( n - k ) p ( 3 t ) = 5 c 3 ( 1 / 2 ) ^ 3 ( 1 / 2 ) ^ 2 = 10 ( 1 / 2 ) ^ 5 p ( 4 t ) = 5 c 4 ( 1 / 2 ) ^ 4 ( 1 / 2 ) ^ 1 = 5 ( 1 / 2 ) ^ 5 = > p ( 3 t ) + p ( 4 t ) = 15 / 32 an... | a ) 5 / 16 , b ) 1 / 2 , c ) 12 / 30 , d ) 15 / 32 , e ) 3 / 8 | d | divide(add(divide(factorial(5), multiply(factorial(const_4), factorial(const_1))), divide(factorial(5), multiply(factorial(const_3), factorial(const_2)))), power(const_2, 5)) | factorial(n1)|factorial(const_4)|factorial(const_1)|factorial(const_3)|factorial(const_2)|power(const_2,n1)|multiply(#1,#2)|multiply(#3,#4)|divide(#0,#6)|divide(#0,#7)|add(#8,#9)|divide(#10,#5) | general |
there are two groups of students in the sixth grade . there are 30 students in group a , and 50 students in group b . if , on a particular day , 20 % of the students in group a forget their homework , and 12 % of the students in group b forget their homework , then what percentage of the sixth graders forgot their home... | total students = 30 + 50 = 80 20 % of 30 = 6 12 & of 50 = 6 total students who forget homework = 6 + 6 = 12 percentage = 1280 ∗ 100 = 15 = 12 / 80 ∗ 100 = 15 ; answer = c = 15 % | a ) 13 % , b ) 14 % , c ) 15 % , d ) 16 % , e ) 17 % | c | multiply(divide(add(divide(multiply(30, 20), const_100), divide(multiply(50, 12), const_100)), add(30, 50)), const_100) | add(n0,n1)|multiply(n0,n2)|multiply(n1,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100) | gain |
a rectangle having length 100 cm and width 40 cm . if the length of the rectangle is increased by fifty percent then how much percent the breadth should be decreased so as to maintain the same area . | "explanation : solution : ( 50 / ( 100 + 50 ) * 100 ) % = 33.33 % answer : b" | a ) 25 % , b ) 33.33 % , c ) 40 % , d ) 75 % , e ) none of these | b | multiply(add(const_1, divide(divide(multiply(100, 40), add(100, divide(multiply(multiply(const_3, const_10), 100), const_100))), 40)), const_10) | multiply(n0,n1)|multiply(const_10,const_3)|multiply(n0,#1)|divide(#2,const_100)|add(n0,#3)|divide(#0,#4)|divide(#5,n1)|add(#6,const_1)|multiply(#7,const_10)| | geometry |
a fair coin is tossed 13 times . what is the probability of getting more heads than tails in 13 tosses ? | "on each toss , the probability of getting a head is 1 / 2 and the probability of getting a tail is 1 / 2 . there is no way to get the same number of heads and tails on an odd number of tosses . there will either be more heads or more tails . then there must be more heads on half of the possible outcomes and more tails... | a ) 1 / 2 , b ) 63 / 128 , c ) 4 / 7 , d ) 61 / 256 , e ) 63 / 64 | a | divide(add(add(add(choose(13, const_2), choose(13, const_3)), choose(13, const_4)), choose(13, 13)), power(const_2, 13)) | choose(n0,const_2)|choose(n0,const_3)|choose(n0,const_4)|choose(n0,n0)|power(const_2,n0)|add(#0,#1)|add(#5,#2)|add(#6,#3)|divide(#7,#4)| | probability |
raman mixed 34 kg of butter at rs . 150 per kg with 36 kg butter at the rate of rs . 125 per kg . at what price per kg should he sell the mixture to make a profit of 40 % in the transaction ? | "explanation : cp per kg of mixture = [ 34 ( 150 ) + 36 ( 125 ) ] / ( 34 + 36 ) = rs . 137.14 sp = cp [ ( 100 + profit % ) / 100 ] = 137.14 * [ ( 100 + 40 ) / 100 ] = rs . 192 . answer : c" | a ) 129 , b ) 287 , c ) 192 , d ) 188 , e ) 112 | c | add(divide(add(multiply(34, 150), multiply(36, 125)), add(36, 34)), multiply(divide(add(multiply(34, 150), multiply(36, 125)), add(36, 34)), divide(40, const_100))) | add(n0,n2)|divide(n4,const_100)|multiply(n0,n1)|multiply(n2,n3)|add(#2,#3)|divide(#4,#0)|multiply(#5,#1)|add(#5,#6)| | gain |
in a simultaneous throw of a pair of dice , find the probability of getting a total more than 5 | "total number of cases = 4 * 4 = 16 favourable cases = [ ( 2,4 ) , ( 3,3 ) , ( 3,4 ) , ( 4,2 ) , ( 4,3 ) , ( 4,4 ) ] = 6 so probability = 6 / 16 = 3 / 8 answer is e" | a ) 1 / 2 , b ) 7 / 12 , c ) 5 / 13 , d ) 5 / 12 , e ) 3 / 8 | e | divide(subtract(5, multiply(const_2, const_3)), 5) | multiply(const_2,const_3)|subtract(n0,#0)|divide(#1,n0)| | general |
in a simultaneous throw of a pair of dice , find the probability of getting a total more than 9 | "total number of cases = 8 * 8 = 64 favourable cases = [ ( 2,8 ) , ( 3,6 ) , ( 3,7 ) , ( 4,6 ) , ( 4,7 ) , ( 4,8 ) , ( 5,5 ) , ( 5,6 ) , ( 5,7 ) , ( 5,8 ) , ( 6,4 ) , ( 6,5 ) , ( 6,6 ) , ( 6,7 ) , ( 6,8 ) , ( 7,3 ) , ( 7,4 ) , ( 7,5 ) , ( 7,6 ) , ( 7,7 ) , ( 7,8 ) , ( 8,2 ) , ( 8,3 ) , ( 8,4 ) , ( 8,5 ) , ( 8,6 ) , ( 8... | a ) 1 / 2 , b ) 7 / 18 , c ) 5 / 13 , d ) 5 / 12 , e ) 6 / 17 | b | divide(subtract(9, multiply(const_2, const_3)), 9) | multiply(const_2,const_3)|subtract(n0,#0)|divide(#1,n0)| | general |
a cubical tank is filled with water to a level of 2 feet . if the water in the tank occupies 32 cubic feet , to what fraction of its capacity is the tank filled with water ? | "the volume of water in the tank is h * l * b = 32 cubic feet . since h = 2 , then l * b = 16 and l = b = 4 . since the tank is cubical , the capacity of the tank is 4 * 4 * 4 = 64 . the ratio of the water in the tank to the capacity is 32 / 64 = 1 / 2 the answer is a ." | a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 3 , d ) 1 / 4 , e ) 3 / 4 | a | divide(2, divide(32, const_10)) | divide(n1,const_10)|divide(n0,#0)| | physics |
a lent rs . 5000 to b for 2 years and rs . 3000 to c for 4 years on simple interest at the same rate of interest and received rs . 2200 in all from both of them as interest . the rate of interest per annum is ? | "let the rate be r % p . a . then , ( 5000 * r * 2 ) / 100 + ( 3000 * r * 4 ) / 100 = 2200 100 r + 120 r = 2200 r = 10 % answer : d" | a ) 16 % , b ) 12 % , c ) 74 % , d ) 10 % , e ) 45 % | d | multiply(divide(2200, add(multiply(5000, 2), multiply(3000, 4))), const_100) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(n4,#2)|multiply(#3,const_100)| | gain |
when 1 / 10 percent of 7000 is subtracted from 1 / 10 of 7000 , the difference is | we can break this problem into two parts : 1 ) what is 1 / 10 percent of 7,000 ? 2 ) what is 1 / 10 of 7,000 ? to calculate 1 / 10 percent of 7,000 we must first remember to divide 1 / 10 by 100 . so we have : ( 1 / 10 ) / ( 100 ) to divide a number by 100 means to multiply it by 1 / 100 , so we have : 1 / 10 x 1 / 100... | a ) 0 , b ) 50 , c ) 450 , d ) 693 , e ) 500 | d | divide(multiply(10, 7000), const_100) | multiply(n1,n2)|divide(#0,const_100) | general |
a train running at the speed of 60 km / hr crosses a pole in 9 sec . what is the length of the train ? | "speed = 60 * 5 / 18 = 50 / 3 m / sec length of the train = speed * time = 50 / 3 * 9 = 150 m answer : e" | a ) 288 , b ) 279 , c ) 277 , d ) 272 , e ) 150 | e | multiply(divide(multiply(60, const_1000), const_3600), 9) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
if 3 x = 9 y = z , what is x + y , in terms of z ? | "3 x = 9 y = z x = z / 3 and y = z / 9 x + y = z / 3 + z / 9 = 4 z / 9 answer is e" | a ) z / 2 , b ) 2 z , c ) z / 3 , d ) 3 z / 5 , e ) 4 z / 9 | e | divide(subtract(divide(multiply(3, const_100), const_2), const_2), add(divide(multiply(3, const_100), const_2), const_2)) | multiply(n0,const_100)|divide(#0,const_2)|add(#1,const_2)|subtract(#1,const_2)|divide(#3,#2)| | general |
into a bag there are 20 honey and 5 cherry candies . if a boy pick only two candies simultaneous and randomly , what is the probability that he picks one candy of each flavor ? | we are told that we have 25 candies , 20 honey and 5 cherry candies . the candies are picked simultaneous and randomly , c 1 and c 2 , in different flavors . there are two acceptable outcomes : 1 ) c 1 is honey and c 2 is cherry ; 2 ) c 1 is cherry and c 2 is honey . let ' s go : 1 ) c 1 = ( 20 / 25 ) ( 5 / 24 ) = 1 / ... | a ) 2 / 25 , b ) 1 / 50 , c ) 1 / 12 , d ) 1 / 6 , e ) 1 / 3 | e | multiply(multiply(divide(20, add(20, 5)), divide(5, add(20, 5))), const_2) | add(n0,n1)|divide(n0,#0)|divide(n1,#0)|multiply(#1,#2)|multiply(#3,const_2) | probability |
if x = 1 + √ 2 , then what is the value of x 4 - 4 x 3 + 4 x 2 + 2 ? | "answer x = 1 + √ 2 ∴ x 4 - 4 x 3 + 4 x 2 + 5 = x 2 ( x 2 - 4 x + 4 ) + 2 = x 2 ( x - 2 ) 2 + 2 = ( 1 + √ 2 ) 2 ( 1 + √ 2 - 2 ) 2 + 2 = ( √ 2 + 1 ) 2 ( √ 2 - 1 ) 2 + 2 = [ ( √ 2 ) 2 - ( 1 ) 2 ] 2 + 2 = ( 2 - 1 ) 2 = 1 + 2 = 3 correct option : e" | a ) - 1 , b ) 0 , c ) 1 , d ) 2 , e ) 3 | e | add(multiply(power(add(1, sqrt(2)), 2), power(subtract(add(1, sqrt(2)), 2), 2)), 4) | sqrt(n1)|add(n0,#0)|power(#1,n1)|subtract(#1,n1)|power(#3,n1)|multiply(#2,#4)|add(n2,#5)| | general |
a line has a slope of 3 / 4 and intersects the point ( - 12 , - 39 ) . at which point e does this line intersect the x - axis ? | assume that the equation of the line is y = mx + c , where m and c are the slope and y - intercept . you are also given that the line crosses the point ( - 12 , - 39 ) , this means that this point will also lie on the line above . thus you get - 39 = m * ( - 12 ) + c , with m = 3 / 4 as the slope is given to be 3 / 4 .... | a ) ( 400 ) , b ) ( 300 ) , c ) ( 040 ) , d ) ( 4030 ) , e ) ( 030 ) | a | multiply(negate(divide(subtract(negate(39), multiply(negate(12), divide(3, 4))), divide(3, 4))), const_10) | divide(n0,n1)|negate(n3)|negate(n2)|multiply(#0,#2)|subtract(#1,#3)|divide(#4,#0)|negate(#5)|multiply(#6,const_10) | general |
the salary of a person was reduced by 35 % . by what percent should his reduced salary be raised so as to bring it at par with his original salary ? | "let the original salary be $ 100 new salary = $ 65 increase on 65 = 35 increase on 100 = 35 / 65 * 100 = 54 % ( approximately ) answer is d" | a ) 50 % , b ) 32 % , c ) 25 % , d ) 54 % , e ) 29 % | d | multiply(divide(multiply(const_100, divide(35, const_100)), subtract(const_100, multiply(const_100, divide(35, const_100)))), const_100) | divide(n0,const_100)|multiply(#0,const_100)|subtract(const_100,#1)|divide(#1,#2)|multiply(#3,const_100)| | gain |
if v and d are both integers , v > d , and - 3 v > 19 , then the largest value of d would be ? | no , your thinking is incorrect . when we know that v > d and v < - 6.33 , the largest value of v can be - 7 while if v = - 7 , then largest value of d < - 7 will be - 8 . for negative numbers , - 7 > - 8 and - 8 > - 10 . you are right in saying that d can take any value less than - 7 - - - > d could be - 8 , - 9 , - 1... | a ) - 5 , b ) - 6 , c ) - 7 , d ) - 8 , e ) - 10 | d | divide(19, 3) | divide(n1,n0) | other |
if ( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 5 ) ( 3 ^ b ) and a and b are positive integers , what is the value of a ? | "( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 5 ) ( 3 ^ b ) = 2 ^ a . 9 ^ a . 9 ^ ( 3 a – 1 ) = ( 2 ^ 5 ) ( 3 ^ b ) just compare powers of 2 from both sides answer = 5 = e" | a ) 22 , b ) 11 , c ) 9 , d ) 6 , e ) 5 | e | multiply(3, 1) | multiply(n2,n3)| | general |
a , b and c invest in the ratio of 3 : 4 : 5 . the percentage of return on their investments are in the ratio of 6 : 5 : 4 . find the total earnings , if b earns rs . 100 more than a : | "explanation : a b c investment 3 x 4 x 5 x rate of return 6 y % 5 y % 4 y % return \ inline \ frac { 18 xy } { 100 } \ inline \ frac { 20 xy } { 100 } \ inline \ frac { 20 xy } { 100 } total = ( 18 + 20 + 20 ) = \ inline \ frac { 58 xy } { 100 } b ' s earnings - a ' s earnings = \ inline \ frac { 2 xy } { 100 } = 100 ... | a ) 2900 , b ) 7250 , c ) 2767 , d ) 1998 , e ) 2771 | a | multiply(add(add(multiply(3, 6), multiply(4, 5)), multiply(5, 4)), divide(100, subtract(multiply(4, 5), multiply(3, 6)))) | multiply(n0,n3)|multiply(n1,n2)|add(#0,#1)|subtract(#1,#0)|add(#2,#1)|divide(n6,#3)|multiply(#4,#5)| | general |
a paint store mixes 3 / 4 pint of red paint and 2 / 3 pint of white paint to make a new paint color called perfect pink . how many pints of red paint would be needed to make 35 pints of perfect pink paint ? | "3 / 4 pint is required to make 3 / 4 + 2 / 3 = 17 / 12 pint of perfect pink so 17 / 12 pint requires 3 / 4 pint of red . . 1 pint will require 3 / 4 * 12 / 17 = 9 / 17 . . 35 pints will require 9 / 17 * 35 = 19 pints . . c" | a ) 9 , b ) 16 , c ) 19 , d ) 25 1 / 3 , e ) 28 1 / 2 | c | multiply(35, divide(multiply(3, 2), multiply(4, 3))) | multiply(n0,n2)|multiply(n0,n1)|divide(#0,#1)|multiply(n4,#2)| | general |
a can build a wall in the same time in which b and c together can do it . if a and b together could do it in 25 days and c alone in 35 days , in what time could b alone do it ? | explanation : no explanation is available for this question ! answer : d | a ) 275 days , b ) 178 days , c ) 185 days , d ) 175 days , e ) 675 days | d | multiply(divide(multiply(25, 35), subtract(35, 25)), const_2) | multiply(n0,n1)|subtract(n1,n0)|divide(#0,#1)|multiply(#2,const_2) | physics |
if 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days , the time taken by 15 men and 20 boys in doing the same type of work will be ? | "let 1 men ' s 1 day work = x and 1 boy ' s 1 day work = y . then , 6 x + 8 y = 1 / 10 and 26 x + 48 y = 1 / 2 solving these two equations , we get : x = 1 / 100 and y = 1 / 200 ( 15 men + 20 boys ) ' s 1 day work = ( 15 / 100 + 20 / 200 ) = 1 / 4 15 men and 20 boys can do the work in 4 days . answer : a" | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | a | divide(multiply(add(6, const_4), 10), add(15, 10)) | add(n0,const_4)|add(n2,n6)|multiply(n2,#0)|divide(#2,#1)| | physics |
a man goes downstream at 11 kmph , and upstream 8 kmph . the speed of the stream is | "speed of the stream = 1 / 2 ( 11 - 8 ) kmph = 1.5 kmph . correct option : a" | a ) 1.5 kmph , b ) 4 kmph , c ) 16 kmph , d ) 2.5 kmph , e ) 26 kmph | a | divide(subtract(11, 8), const_2) | subtract(n0,n1)|divide(#0,const_2)| | physics |
in a 1000 m race , a beats b by 70 m and b beats c by 100 m . in the same race , by how many meters does a beat c ? | by the time a covers 1000 m , b covers ( 1000 - 70 ) = 930 m . by the time b covers 1000 m , c covers ( 1000 - 100 ) = 900 m . so , the ratio of speeds of a and c = 1000 / 930 * 1000 / 900 = 1000 / 837 so , by the time a covers 1000 m , c covers 837 m . so in 1000 m race a beats c by 1000 - 837 = 163 m . answer : d | a ) 145 m , b ) 176 m , c ) 168 m , d ) 163 m , e ) 218 m | d | subtract(1000, divide(multiply(subtract(1000, 70), subtract(1000, 100)), 1000)) | subtract(n0,n1)|subtract(n0,n2)|multiply(#0,#1)|divide(#2,n0)|subtract(n0,#3) | physics |
what is the sum of the multiples of 4 from 40 to 80 , inclusive ? | "the formula we want to use in this type of problem is this : average * total numbers = sum first , find the average by taking the sum of the f + l number and divide it by 2 : a = ( f + l ) / 2 second , find the total numbers in our range by dividing our f and l numbers by 4 and add 1 . ( 80 / 4 ) - ( 40 / 4 ) + 1 mult... | a ) 560 , b ) 660 , c ) 800 , d ) 760 , e ) 480 | b | multiply(divide(add(subtract(80, const_3), add(40, const_2)), const_2), add(divide(subtract(subtract(80, const_3), add(40, const_2)), 4), const_1)) | add(n1,const_2)|subtract(n2,const_3)|add(#0,#1)|subtract(#1,#0)|divide(#3,n0)|divide(#2,const_2)|add(#4,const_1)|multiply(#6,#5)| | general |
what number times ( 1 ⁄ 2 ) ^ 2 will give the value of 2 ^ 3 ? | x * ( 1 / 2 ) ^ 2 = 2 ^ 3 x = 2 ^ 2 * 2 ^ 3 = 2 ^ 5 = 32 the answer is e . | a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) 32 | e | multiply(power(2, 2), power(2, 3)) | power(n1,n1)|power(n1,n4)|multiply(#0,#1) | general |
a bowl was filled with 10 ounces of water , and 0.00008 ounce of the water evaporated each day during a 50 - day period . what percent of the original amount of water evaporated during this period ? | "total amount of water evaporated each day during a 50 - day period = . 00008 * 50 = . 00008 * 100 / 2 = . 008 / 2 = . 004 percent of the original amount of water evaporated during this period = ( . 004 / 10 ) * 100 % = 0.04 % answer b" | a ) 0.004 % , b ) 0.04 % , c ) 0.40 % , d ) 4 % , e ) 40 % | b | multiply(divide(multiply(50, 0.00008), 10), const_100) | multiply(n1,n2)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a car travelling with 2 / 3 km of its actual speed covers 12 km in 2 hr 14 min 28 sec find the actual speed of the car ? | "time taken = 2 hr 14 min 28 sec = 823 / 50 hrs let the actual speed be x kmph then 2 / 3 x * 823 / 50 = 12 x = = 1.09 kmph answer ( c )" | a ) 8.9 kmph , b ) 2.96 kmph , c ) 1.09 kmph , d ) 45.9 kmph , e ) 4.8 kmph | c | divide(multiply(divide(12, add(2, divide(add(multiply(14, const_60), 28), const_3600))), 3), 2) | multiply(n4,const_60)|add(n5,#0)|divide(#1,const_3600)|add(n3,#2)|divide(n2,#3)|multiply(n1,#4)|divide(#5,n0)| | physics |
if teena is driving at 55 miles per hour and is currently 7.5 miles behind joe , who is driving at 40 miles per hour in the same direction then in how many minutes will teena be 30 miles ahead of joe ? | "this type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30 - 40 seconds for a difficult one . teena covers 55 miles in 60 mins . joe covers 40 miles in 60 mins so teena gains 15 miles every 60 mins teena need to cover 7.5 + 30 miles . teena ca... | a ) 150 , b ) 60 , c ) 75 , d ) 90 , e ) 105 | a | multiply(divide(add(subtract(55, 40), 7.5), subtract(55, 40)), const_60) | subtract(n0,n2)|add(n1,#0)|divide(#1,#0)|multiply(#2,const_60)| | physics |
the least number which when increased by 4 each divisible by each one of 22 , 32 , 36 and 54 is : | "solution required number = ( l . c . m . of 24 , 32 , 36 , 54 ) - 4 = 864 - 4 = 860 . answer c" | a ) 427 , b ) 859 , c ) 860 , d ) 4320 , e ) none of these | c | subtract(lcm(lcm(lcm(22, 32), 36), 54), 4) | lcm(n1,n2)|lcm(n3,#0)|lcm(n4,#1)|subtract(#2,n0)| | general |
a trader sells 30 meters of cloth for rs . 4500 at the profit of rs . 10 per metre of cloth . what is the cost price of one metre of cloth ? | "sp of 1 m of cloth = 4500 / 30 = rs . 150 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 150 - rs . 10 = rs . 140 . answer : c" | a ) rs . 80 , b ) rs . 185 , c ) rs . 140 , d ) rs . 295 , e ) none of these | c | subtract(divide(4500, 30), 10) | divide(n1,n0)|subtract(#0,n2)| | physics |
find the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 10 cm . | "explanation : area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 10 ) = 190 cm 2 answer : option b" | a ) 287 cm 2 , b ) 190 cm 2 , c ) 180 cm 2 , d ) 785 cm 2 , e ) 295 cm 2 | b | quadrilateral_area(10, 18, 20) | quadrilateral_area(n2,n1,n0)| | physics |
if 35 percent of 400 is 20 percent of x , then what is 70 percent of x ? | "35 / 100 ( 400 ) = 2 / 10 ( x ) x = 700 . . 70 percent of x = 70 / 100 ( 700 ) = 490 option b ." | a ) 200 , b ) 490 , c ) 700 , d ) 900 , e ) 1,400 | b | divide(multiply(35, 400), 20) | multiply(n0,n1)|divide(#0,n2)| | gain |
for any even integer p , 300 multiplied by p is square of an integer . what is the least value of p ? | p ∗ 3 ∗ 10 ^ 2 = s ^ 2 so , s = √ p ∗ 3 ∗ 10 ^ 2 so , if p = 3 we get a perfect square number ! ! hence answer will be ( a ) 3 | a ) 3 , b ) 4 , c ) 10 , d ) 12 , e ) 14 | a | divide(300, const_100) | divide(n0,const_100) | geometry |
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