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a group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 7 with 3 students left over . what is the sum of the two smallest possible values of n ? | "a 410 a = ( 72 * 5 / 18 ) * 28 - 150 = 410" | a ) 410 m , b ) 354 m , c ) 450 m , d ) 350 m , e ) 250 m | a | subtract(multiply(28, multiply(72, const_0_2778)), 150) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
a train 150 m long running at 72 kmph crosses a platform in 28 sec . what is the length of the platform ? | "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 Γ’ β¬ β 2552 = 208 runs . let the highest score be x , hence the lowest score = x Γ’ β¬ β 170 x + ( x - 170 ) = 208 2 x = 378 x = 189 runs answer : c" | a ) 179 , b ) 367 , c ) 189 , d ) 177 , e ) 191 | c | divide(add(170, 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 batting average of a particular batsman is 60 runs in 46 innings . if the difference in his highest and lowest score is 170 runs and his average excluding these two innings is 58 runs , find his highest score . | "x 13 : ( t , d , s ) y 14 : ( t , 3 d , s + 1000 mi / hour ) z 15 : ( t + 2 seconds , s , 5 d ) d = ? distance = speed * time x 13 : d = s * t x 14 : 3 d = ( s + 1000 ) * t = = = > 3 d = ts + 1000 t z 15 : 5 d = s * ( t + 2 t ) = = = > 5 d = st + 2 st = = = > 5 d - 2 st = st 3 d = 5 d - 2 st + 1000 t - 2 d = - 2 st + 1000 t 2 d = 2 st - 1000 t d = st - 500 t x 13 : d = s * t st - 500 t = s * t s - 500 = s - 250 = s i got to this point and could n ' t go any further . this seems like a problem where i can set up individual d = r * t formulas and solve but it appears that ' s not the case . for future reference how would i know not to waste my time setting up this problem in the aforementioned way ? thanks ! ! ! the distance of z 15 is equal to five times the distance of x 13 ( we established that x 13 is the baseline and thus , it ' s measurements are d , s , t ) s ( t + 2 ) = 5 ( s * t ) what clues would i have to know to set up the equation in this fashion ? is it because i am better off setting two identical distances together ? st + 2 s = 5 st t + 2 = 5 t 2 = 4 t t = 1 / 2 we are looking for distance ( d = s * t ) so we need to solve for speed now that we have time . speed y 14 - speed x 13 speed = d / t 3 d / t - d / t = 1000 ( remember , t is the same because both asteroids were observed for the same amount of time ) 2 d = 1000 2 = 500 d = s * t d = 500 * ( 1 / 2 ) d = 250 answer : a" | a ) 250 , b ) 1,600 / 3 , c ) 1,000 , d ) 1,500 , e ) 2,500 | a | multiply(divide(1000, 2), divide(const_1, 2)) | divide(n8,n3)|divide(const_1,n3)|multiply(#0,#1)| | physics |
the speeds of three asteroids were compared . asteroids x - 13 and y - 14 were observed for identical durations , while asteroid z - 15 was observed for 2 seconds longer . during its period of observation , asteroid y - 14 traveled three times the distance x - 13 traveled , and therefore y - 14 was found to be faster than x - 13 by 1000 kilometers per second . asteroid z - 15 had an identical speed as that of x - 13 , but because z - 15 was observed for a longer period , it traveled five times the distance x - 13 traveled during x - 13 ' s inspection . asteroid x - 13 traveled how many kilometers during its observation ? | "we are asked to find the percentage of females in employed people . total employed people 64 % , out of which 35 are employed males , hence 29 % are employed females . ( employed females ) / ( total employed people ) = 29 / 64 = 45 % answer : d ." | a ) 16 % , b ) 25 % , c ) 32 % , d ) 45 % , e ) 52 % | d | multiply(divide(subtract(64, 35), 64), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
in town x , 64 percent of the population are employed , and 35 percent of the population are employed males . what percent of the employed people in town x are females ? | "in a room of n people , the number of possible handshakes is c ( n , 2 ) or n ( n - 1 ) / 2 so n ( n - 1 ) / 2 = 91 or n ( n - 1 ) = 182 or n = 14 answer is ( a )" | a ) 14 , b ) 12 , c ) 11 , d ) 15 , e ) 16 | a | divide(divide(multiply(91, const_2), const_3), const_4) | multiply(n0,const_2)|divide(#0,const_3)|divide(#1,const_4)| | general |
everyone shakes hands with everyone else in a room . total number of handshakes is 91 . number of persons = ? | "answer given expression = 12008 - 50 Γ· 10.00 = 12008 - 5 = 12003 correct option : d" | a ) 1195 , b ) 120 , c ) 12000 , d ) 12003 , e ) none | d | subtract(multiply(divide(12008, const_100), 50), multiply(divide(const_1, const_3), multiply(divide(12008, const_100), 50))) | divide(n0,const_100)|divide(const_1,const_3)|multiply(n1,#0)|multiply(#1,#2)|subtract(#2,#3)| | general |
12008 - 50 Γ· 10.00 = ? | "we can write the above in terms of ( a + b ) ( a - b ) 6 ^ 4 β 4 ^ 4 = ( 6 ^ 2 ) 2 - ( 4 ^ 2 ) 2 = ( 6 ^ 2 β 4 ^ 2 ) * ( 6 ^ 2 + 4 ^ 2 ) = ( 36 β 16 ) * ( 36 + 16 ) = > 20 * 52 = 1040 ans option e ." | a ) 20 , b ) 52 , c ) 104 , d ) 520 , e ) 1040 | e | divide(power(6, 4), power(6, 4)) | power(n0,n1)|power(n0,n3)|divide(#0,#1)| | general |
6 ^ 4 β 4 ^ 4 = ? | "( 855000 + 220 ) / 900 * 900 = 855000 + 220 = 855220 answer : d" | a ) 854542 , b ) 856945 , c ) 758965 , d ) 855220 , e ) 826450 | d | multiply(add(divide(220, 900), 950), 900) | divide(n1,n2)|add(n0,#0)|multiply(#1,n2)| | general |
find the value of ( 950 + 220 / 900 ) Γ 900 | if a + b = 4.6 assuming a and b are positive then a * b < 10 ( they could be either 1,2 , 3,4 ) therefore a * b = c 2 * b = c a + b = 4.6 three simple equations - divide the 1 st / 2 nd - - > a = 2 plug it the 3 rd - - > b = 2.6 - - > c = 5.2 ( answer b ) | a ) 6 , b ) 5.2 , c ) 4 , d ) 3 , e ) 2 | b | divide(multiply(subtract(4.6, 2), const_10), add(const_4, const_1)) | add(const_1,const_4)|subtract(n1,n0)|multiply(#1,const_10)|divide(#2,#0) | general |
__ 2 a x __ b ____ cc in the multiplication problem above , a , b , and c represent distinct digits . if the sum of a and b is equal to 4.6 , what is the value of c ? | let t be the tier price , p be total price = 18000 per the given conditions : 0.12 t + 0.09 ( p - t ) = 1950 0.12 t + 0.09 * 18000 - 0.09 t = 1950 0.03 t + 1620 = 1950 0.03 t = 1950 - 1620 = 330 t = 330 / 0.03 = 11000 answer b | a ) $ 11500 , b ) $ 11000 , c ) $ 12000 , d ) $ 12100 , e ) $ 12500 | b | divide(subtract(1950, multiply(multiply(multiply(const_3, multiply(const_2, const_3)), const_1000), divide(9, const_100))), subtract(divide(12, const_100), divide(9, const_100))) | divide(n1,const_100)|divide(n0,const_100)|multiply(const_2,const_3)|multiply(#2,const_3)|subtract(#1,#0)|multiply(#3,const_1000)|multiply(#0,#5)|subtract(n3,#6)|divide(#7,#4) | general |
country c imposes a two - tiered tax on imported cars : the first tier imposes a tax of 12 % of the car ' s price up to a certain price level . if the car ' s price is higher than the first tier ' s level , the tax on the portion of the price that exceeds this value is 9 % . if ron imported a $ 18,000 imported car and ended up paying $ 1950 in taxes , what is the first tier ' s price level ? | "( 4 - x ) = x * ( 3 + x ) ( 4 - x ) = 3 x + x ^ 2 0 = x ^ 2 + 4 x - 4 the answer is c ." | a ) - 4 , b ) - 1 , c ) 0 , d ) 1 , e ) 4 | c | subtract(multiply(4, 2), 4) | multiply(n2,n0)|subtract(#0,n0)| | general |
if ( 4 - x ) / ( 3 + x ) = x , what is the value of x ^ 2 + 4 x - 4 ? | "by graphing the points , we can see that this figure is a trapezoid . a trapezoid is any quadrilateral that has one set of parallel sides , and the formula for the area of a trapezoid is : area = ( 1 / 2 ) Γ ( base 1 + base 2 ) Γ ( height ) , where the bases are the parallel sides . we can now determine the area of the quadrilateral : area = 1 / 2 Γ ( 8 + 7 ) Γ 12 = 90 . the answer is a ." | a ) 90 , b ) 95 , c ) 104 , d ) 117 , e ) 182 | a | divide(multiply(add(add(2, 6), add(2, 6)), subtract(14, 2)), 2) | add(n1,n3)|add(n3,n5)|subtract(n4,n0)|add(#0,#1)|multiply(#3,#2)|divide(#4,n5)| | geometry |
in a rectangular coordinate system , what is the area of a quadrilateral whose vertices have the coordinates ( 2 , - 2 ) , ( 2 , 6 ) , ( 14 , 2 ) , ( 14 , - 5 ) ? | "time taken to reach = 3000 / 40 = 75 minutes answer : c" | a ) 70 minutes , b ) 72 minutes , c ) 75 minutes , d ) 76 minutes , e ) 77 minutes | c | divide(3000, 40) | divide(n1,n0)| | gain |
a scuba diver descends at a rate of 40 feet per minute . a diver dive from a ship to search for a lost ship at the depth of 3000 feet below the sea level . . how long will he take to reach the ship ? | "explanation : ( 1 man + 1 woman + 1 boy ) β s 1 day β s work = 1 / 3 1 man β s 1 day work = 1 / 6 1 boy β s 1 day β s work = 1 / 18 ( 1 man + 1 boy ) β s 1 day β s work = 1 / 6 + 1 / 18 = 2 / 9 therefore , 1 woman β s 1 day β s work = 1 / 3 β 2 / 9 = 3 - 2 / 9 = 1 / 9 therefore , the woman alone can finish the work in 9 days . answer : option a" | a ) 9 days , b ) 21 days , c ) 24 days , d ) 27 days , e ) 28 days | a | inverse(subtract(inverse(3), add(inverse(6), inverse(18)))) | inverse(n0)|inverse(n1)|inverse(n2)|add(#1,#2)|subtract(#0,#3)|inverse(#4)| | physics |
a man , a woman and a boy can together complete a piece of work in 3 days . if a man alone can do it in 6 days and a boy alone in 18 days , how long will a woman take to complete the work ? | "explanation : let the required number of bottles be x . more machines , more bottles ( direct proportion ) more minutes , more bottles ( direct proportion ) machines 6 : 10 | | : : 270 : x time 1 : 4 | = > 6 * 1 * x = 10 * 4 * 270 = > x = ( 10 * 4 * 270 ) / 6 = > x = 1800 answer : d" | a ) 1500 , b ) 1545.6 , c ) 1640.33 , d ) 1800 , e ) none of these | d | multiply(multiply(divide(270, 6), 4), 10) | divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)| | gain |
running at the same constant rate , 6 identical machines can produce a total of 270 pens per minute . at this rate , how many pens could 10 such machines produce in 4 minutes ? | "the tank is emptied at this rate : 14 + 6 - 5 = 15 cubic inches / min the tank has a volume of 45 * 12 * 12 * 12 = 77760 cubic inches . the time it takes to empty the tank is 77760 / 15 = 5184 minutes . the answer is a ." | a ) 5184 , b ) 5276 , c ) 5368 , d ) 5460 , e ) 5552 | a | divide(multiply(45, power(14, 5)), subtract(add(12, 6), 5)) | add(n4,n5)|power(n8,n6)|multiply(n0,#1)|subtract(#0,n2)|divide(#2,#3)| | physics |
a tank with a volume of 45 cubic feet has one inlet pipe and 2 outlet pipes . the inlet pipe fills water into the tank at the rate of 5 cubic inches / min and the 2 outlet pipes empty it out at the rates of 14 cubic inches / min and 6 cubic inches / min respectively . if all 3 pipes are opened when the tank is full , how many minutes does it take to empty the tank ? ( 1 foot = 12 inches ) | "cost price * 1.13 = selling price - - > cost price * 1.13 = $ 28 - - > cost price = $ 24.78 . answer : c ." | a ) 3.0 , b ) 3.36 , c ) 24.78 , d ) 25.0 , e ) 31.36 | c | divide(multiply(28, const_100), add(const_100, 13)) | add(n0,const_100)|multiply(n1,const_100)|divide(#1,#0)| | gain |
from the sale of sleeping bags , a retailer made a gross profit of 13 % of the wholesale cost . if each sleeping bag was sold for $ 28 , what was the wholesale cost per bag ? | final number = initial number + 18 % ( original number ) = 86 + 18 % ( 86 ) = 86 + 15 = 101 . answer e | a ) 105 , b ) 95 , c ) 80 , d ) 60 , e ) 101 | e | multiply(86, add(const_1, divide(18, const_100))) | divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1) | gain |
a light has a rating of 86 watts , it is replaced with a new light that has 18 % higher wattage . how many watts does the new light have ? | let the man ' s rate upstream be x kmph and that downstream be y kmph . then , distance covered upstream in 8 hrs 48 min = distance covered downstream in 4 hrs . 44 * x / 5 = 4 * y y = 11 / 5 * x required ratio = ( y + x ) / 2 : ( y - x ) / 2 = 8 / 5 : 3 / 5 = 8 / 3 ans - b | a ) 8 / 5 , b ) 8 / 3 , c ) 3 / 5 , d ) 5 / 8 , e ) 5 / 3 | b | divide(divide(add(divide(add(divide(48, const_60), 8), 4), const_1), const_2), divide(subtract(divide(add(divide(48, const_60), 8), 4), const_1), const_2)) | divide(n1,const_60)|add(n0,#0)|divide(#1,n2)|add(#2,const_1)|subtract(#2,const_1)|divide(#3,const_2)|divide(#4,const_2)|divide(#5,#6) | physics |
a boat running upstream takes 8 hours 48 minutes to cover a certain distance , while it takes 4 hours to cover the same distance running downstream . what is the ratio between the speed of the boat and speed of the water current respectively ? | "no . of toys = 100 / 1 = 100 answer : e" | a ) 40 , b ) 54 , c ) 45 , d ) 39 , e ) 100 | e | divide(100, 1) | divide(n1,n0)| | physics |
a worker makes a toy in every 1 h . if he works for 100 h , then how many toys will he make ? | "20 % of 3000 gives 600 . so 600 attends chess and 40 % of 600 gives 240 so 240 enrolled for swimming answer : a" | a ) 240 , b ) 10 , c ) 100 , d ) 50 , e ) 20 | a | divide(multiply(divide(multiply(20, 3000), const_100), 40), const_100) | multiply(n0,n1)|divide(#0,const_100)|multiply(n2,#1)|divide(#2,const_100)| | gain |
there are 3000 students in a school and among them 20 % of them attends chess class . 40 % of the students who are in the chess class are also enrolled for swimming . no other students are interested in swimming so how many will attend the swimming class if all enrolled attends ? | "p = 280 r = 10 % required population of town = p * ( 1 + r / 100 ) ^ t = 280 * ( 1 + 10 / 100 ) = 280 * ( 11 / 10 ) = 308 answer is e" | a ) 100 , b ) 120 , c ) 200 , d ) 220 , e ) 308 | e | add(280, divide(multiply(280, 10), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)| | gain |
the present population of a town is 280 . population increase rate is 10 % p . a . find the population of town after 1 years ? | # of boxes of cookies mary sold = x ann sold 5 times more = 5 x x + 5 x = 18 6 x = 18 x = 18 / 6 = 3 answer : a | a ) 3 , b ) 5 , c ) 6 , d ) 10 , e ) 18 | a | divide(18, add(5, const_1)) | add(n0,const_1)|divide(n1,#0) | general |
mary sold boxes of butter cookies . ann sold 5 times as much as she did . 18 boxes of cookies were sold that day , how many boxes did mary sell ? | 8 x + 75 = 9 ( x + 7 ) x = 12 + 7 = 19 answer : b | a ) 12 , b ) 19 , c ) 26 , d ) 33 , e ) 40 | b | add(subtract(75, multiply(7, 9)), 7) | multiply(n0,n2)|subtract(n1,#0)|add(n2,#1) | general |
a batsman in his 9 th inning makes a score of 75 and their by increasing his average by 7 . what is his average after the 9 th inning ? | "explanation : speed in still water = 5 kmph speed of the current = 1 kmph speed downstream = ( 5 + 1 ) = 6 kmph speed upstream = ( 5 - 1 ) = 4 kmph let the requited distance be x km total time taken = 1 hour = > x / 6 + x / 4 = 1 = > 2 x + 3 x = 12 = > 5 x = 12 = > x = 2.4 km . answer : option c" | a ) 3.2 km , b ) 3 km , c ) 2.4 km , d ) 3.6 km , e ) none of these | c | divide(multiply(subtract(5, 1), const_3), 5) | subtract(n0,n1)|multiply(#0,const_3)|divide(#1,n0)| | physics |
a man can row at 5 kmph in still water . if the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back , how far is the place ? | rate = work / time given rate of machine a = x / 10 min machine b produces 2 x boxes in 5 min hence , machine b produces 4 x boxes in 10 min . rate of machine b = 4 x / 10 we need tofind the combined time that machines a and b , working simultaneouslytakeat their respective constant rates let ' s first find the combined rate of machine a and b rate of machine a = x / 10 min + rate of machine b = 4 x / 10 = 5 x / 10 now combine time = combine work needs to be done / combine rate = 4 x / 5 x * 10 = 8 min ans : e | a ) 3 minutes , b ) 4 minutes , c ) 5 minutes , d ) 6 minutes , e ) 8 minutes | e | divide(multiply(4, 10), add(speed(10, 10), speed(multiply(2, 10), 5))) | multiply(n0,n3)|multiply(n0,n1)|speed(n0,n0)|speed(#1,n2)|add(#2,#3)|divide(#0,#4) | physics |
working alone at its constant rate , machine a produces x boxes in 10 minutes and working alone at its constant rate , machine b produces 2 x boxes in 5 minutes . how many minutes does it take machines a and b , working simultaneously at their respective constant rates , to produce 4 x boxes ? | "the bottom row has x bricks x + x - 1 + x - 2 + x - 3 = 134 4 x - 6 = 134 4 x = 128 x = 32 answer : c" | a ) 30 , b ) 31 , c ) 32 , d ) 33 , e ) 34 | c | divide(subtract(subtract(subtract(subtract(134, const_1), const_2), const_3), const_4), 4) | subtract(n1,const_1)|subtract(#0,const_2)|subtract(#1,const_3)|subtract(#2,const_4)|divide(#3,n0)| | general |
in a certain brick wall , each row of bricks above the bottom row contains one less brick than the row just below it . if there are 4 rows in all and a total of 134 bricks in the wall , how many bricks does the bottom row contain ? | "total surface area = 6 a ^ 2 = 6 * 10 * 10 = 600 each quart covers 20 sqr ft thus total number of quarts = 600 / 20 = 30 cost will be 30 * 3.1 = $ 93 ans : c" | a ) $ 1.60 , b ) $ 16.00 , c ) $ 93.00 , d ) $ 108.00 , e ) $ 196.00 | c | multiply(divide(3.10, 20), surface_cube(10)) | divide(n0,n1)|surface_cube(n2)|multiply(#0,#1)| | geometry |
if paint costs $ 3.10 per quart , and a quart covers 20 square feet , how much will it cost to paint the outside of a cube 10 feet on each edge ? | "9 / 12 = 3 / 4 * 1000 = 750 550 - - - - - - - - - - - - - 200 1 / 4 - - - - - - - - 200 1 - - - - - - - - - ? = > rs . 800 answer : d" | a ) s . 80 , b ) s . 85 , c ) s . 90 , d ) s . 800 , e ) s . 120 | d | multiply(divide(subtract(multiply(9, 1000), multiply(multiply(const_3, const_4), 550)), multiply(multiply(const_3, const_4), const_1)), const_4) | multiply(n0,n1)|multiply(const_3,const_4)|multiply(n2,#1)|multiply(#1,const_1)|subtract(#0,#2)|divide(#4,#3)|multiply(#5,const_4)| | general |
a man engaged a servant on the condition that he would pay him rs . 1000 and a uniform after one year service . he served only for 9 months and received uniform and rs . 550 , find the price of the uniform ? | start with the prime factorization : 33150 = 50 * 663 = ( 2 * 5 * 5 ) * 3 * 221 = ( 2 ) * ( 3 ) * ( 5 ^ 2 ) * ( 13 ) * ( 17 ) there are five distinct prime factors , { 2 , 3 , 5 , 13 , and 17 } answer : b . | a ) four , b ) five , c ) six , d ) seven , e ) eight | b | add(const_2, const_3) | add(const_2,const_3) | other |
how many distinct prime numbers are factors of 33150 ? | "in 12 days 12 * 0.06 = 0.72 ounces of water evaporated , which is 0.72 / 24 Γ’ Λ β 100 = 3 of the original amount of water . answer : c ." | a ) 0.003 % , b ) 0.03 % , c ) 3 % , d ) 2 % , e ) 30 % | c | multiply(divide(multiply(0.06, 12), 24), const_100) | multiply(n1,n2)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a glass was filled with 24 ounces of water , and 0.06 ounce of the water evaporated each day during a 12 - day period . what percent of the original amount of water evaporated during this period ? | solution : let x be the number i chose , then x / 4 Γ’ Λ β 18 = 7 x / 4 = 25 x = 100 answer b | a ) 600 , b ) 100 , c ) 800 , d ) 900 , e ) none | b | multiply(add(18, 7), 4) | add(n1,n2)|multiply(n0,#0) | general |
i chose a number and divide it by 4 . then i subtracted 18 from the result and got 7 . what was the number i chose ? | let ' a ' can be read hindu , let ' b ' can be read times , let ' c ' can be read deccan , from the given data : n ( aubuc ) = 76 , n ( a ) = 54 , n ( b ) = 43 , n ( c ) = 37 , n ( anbnc ) = 15 , n ( anc ) = 24 , n ( bnc ) = 27 , n ( anb ) = ? n ( aubuc ) = n ( a ) + n ( b ) + n ( c ) - n ( anb ) - n ( bnc ) - n ( anc ) + n ( anbnc ) = = > 76 = 54 + 43 + 37 - n ( anb ) - 24 - 27 + 15 = = > n ( anb ) = 54 + 43 + 37 + 15 - 24 - 27 - 76 = = > n ( anb ) = 149 - 127 = = > n ( anb ) = 22 answer : b | a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | b | add(subtract(24, 15), subtract(27, 15)) | subtract(n5,n4)|subtract(n6,n4)|add(#0,#1) | general |
there are 76 persons . 54 can read hindu , 43 can read times , 37 can read deccan and 15 can read all . if 24 can read hindu and deccan and 27 can read deccan and times then what is the number of persons who read only times and hindu . | "0.8 * ( 1.2 * 100 ) = $ 96 the answer is b ." | a ) $ 88 , b ) $ 96 , c ) $ 100 , d ) $ 106 , e ) $ 110 | b | subtract(add(100, divide(multiply(100, 20), const_100)), divide(multiply(add(100, divide(multiply(100, 20), const_100)), 20), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|multiply(n1,#2)|divide(#3,const_100)|subtract(#2,#4)| | general |
the original price of a suit is $ 100 . the price increased 20 % , and after this increase , the store published a 20 % off coupon for a one - day sale . given that the consumers who used the coupon on sale day were getting 20 % off the increased price , how much did these consumers pay for the suit ? | "total score of the batsman in 20 matches = 800 . total score of the batsman in the next 30 matches = 600 . total score of the batsman in the 50 matches = 1400 . average score of the batsman = 1400 / 50 = 28 . answer : c" | a ) 31 , b ) 46 , c ) 28 , d ) 13 , e ) 12 | c | divide(add(multiply(40, 20), multiply(20, 30)), add(20, 30)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)| | general |
the average runs scored by a batsman in 20 matches is 40 . in the next 30 matches the batsman scored an average of 20 runs . find his average in all the 50 matches ? | "let x be the first integer in the set , then x + 54 is the largest integer . the sum is : x + ( x + 1 ) + ( x + 2 ) + . . . + ( x + 54 ) = 55 x + 54 * 55 / 2 = 55 ( x + 27 ) then x + 27 = 101 x = 74 the largest integer in the set is 74 + 54 = 128 the answer is b ." | a ) 126 , b ) 128 , c ) 130 , d ) 132 , e ) 134 | b | add(add(power(add(add(divide(subtract(subtract(55, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(55, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(55, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(55, const_10), const_2), const_4), const_2), const_2))) | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics |
the sum of 55 consecutive integers is 5555 . what is the greatest integer in the set ? | "a number of valid votes = 80 % of 5000 = 4000 . valid votes polled by other candidate = 45 % of 4000 = ( 45 / 100 x 4000 ) = 1800 ." | a ) 1800 , b ) 2700 , c ) 2900 , d ) 2200 , e ) 2300 | a | multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 5000) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(n2,#4)| | gain |
in an election between two candidates , one got 55 % of the total valid votes , 20 % of the votes were invalid . if the total number of votes was 5000 , the number of valid votes that the other candidate got , was : | "let x = ounces of 60 % salt solution to be added . 2 * 50 + . 6 x = . 5 ( 50 + x ) x = 150 answer e" | a ) 16.67 , b ) 30 , c ) 50 , d ) 60.33 , e ) 150 | e | divide(subtract(multiply(divide(50, const_100), 50), multiply(divide(20, const_100), 50)), subtract(divide(50, const_100), divide(20, const_100))) | divide(n3,const_100)|divide(n2,const_100)|multiply(n1,#0)|multiply(n1,#1)|subtract(#0,#1)|subtract(#2,#3)|divide(#5,#4)| | gain |
how many ounces of a 60 % salt solution must be added to 50 ounces of a 20 percent salt solution so that the resulting mixture is 50 % salt ? | index for females = ( 20 - 8 ) / 20 = 3 / 5 = 0.6 index for males = ( 20 - 12 / 20 = 2 / 5 = 0.4 index for females exceeds males by 0.6 - 0.4 = 0.2 answer : c | a ) 0.05 , b ) 0.0625 , c ) 0.2 , d ) 0.25 , e ) 0.6 | c | subtract(divide(subtract(20, 8), 20), divide(8, 20)) | divide(n1,n0)|subtract(n0,n1)|divide(#1,n0)|subtract(#2,#0) | general |
for a group of n people , k of whom are of the same sex , the ( n - k ) / n expression yields an index for a certain phenomenon in group dynamics for members of that sex . for a group that consists of 20 people , 8 of whom are females , by how much does the index for the females exceed the index for the males in the group ? | let number of packs of four = f let number of packs of nine = n 4 f + 9 n = 97 now , we need to test for values of n . since sum 97 is odd and 4 f will always be even , n ca n ' t be even . now , we can test for values e = 2 , 4 and 6 4 * 4 + 9 * 9 = 16 + 81 = 97 answer d | a ) 3 , b ) 4 , c ) 8 , d ) 9 , e ) 13 | d | divide(subtract(97, multiply(4, 4)), 9) | multiply(n0,n0)|subtract(n2,#0)|divide(#1,n1) | general |
chocolate bars are sold in packages of 4 or 9 only . if mark bought 97 chocolate bars exactly , what could be the number of large packs mark bought ? | "let c . p . be rs . x . then , ( 1920 - x ) / x * 100 = ( x - 1280 ) / x * 100 1920 - x = x - 1280 2 x = 3200 = > x = 1600 required s . p . = 120 % of rs . 1600 = 120 / 100 * 1600 = rs . 1920 . answer : d" | a ) 2000 , b ) 2778 , c ) 2299 , d ) 1920 , e ) 2771 | d | multiply(divide(add(const_100, 20), const_100), divide(add(1920, 1280), const_2)) | add(n2,const_100)|add(n0,n1)|divide(#0,const_100)|divide(#1,const_2)|multiply(#2,#3)| | gain |
the percentage profit earned by selling an article for rs . 1920 is equal to the percentage loss incurred by selling the same article for rs . 1280 . at what price should the article be sold to make 20 % profit ? | "the time spent on section b and section c is 27 minutes each . the ratio of c to a is 27 : 6 = 9 : 2 the answer is b ." | a ) 7 : 1 , b ) 9 : 2 , c ) 8 : 3 , d ) 5 : 2 , e ) 6 : 1 | b | divide(3, const_4) | divide(n0,const_4)| | physics |
a student completes a 3 part exam as follows . the student completes sections a in 6 minutes and takes an equal time to complete the two remaining sections . what is the ratio of time spent on section c to section a if the total duration is 1 hour ? | "on dividing 531742 by 3 we get the remainder 1 , so 1 should be subtracted c" | a ) 4 , b ) 5 , c ) 1 , d ) 2 , e ) 3 | c | subtract(531742, multiply(floor(divide(531742, 3)), 3)) | divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)| | general |
find the least number must be subtracted from 531742 so that remaining no . is divisible by 3 ? | 10 men - working 2 hrs - cut 10 trees 1 men - working 1 hr - cuts = 10 / 10 * 2 thus 8 men - working 3 hrs - cut = 10 * 8 * 3 / 10 * 2 = 12 trees answer is a | a ) 12 , b ) 15 , c ) 16 , d ) 18 , e ) 20 | a | multiply(multiply(subtract(10, 2), divide(divide(10, 2), 10)), 3) | divide(n0,n2)|subtract(n0,n2)|divide(#0,n0)|multiply(#2,#1)|multiply(n4,#3) | physics |
10 men can cut 10 trees in 2 hours . if 2 men leave the job , how many trees will be cut in 3 hours ? | "let the mother ' s present age be x years then the person ' s present age = 2 x / 5 ( 3 x / 5 ) + 8 = 1 / 2 ( x + 8 ) 2 ( 3 x + 40 ) = 5 ( x + 8 ) x = 40 answer is e" | a ) a ) 25 , b ) b ) 44 , c ) c ) 32 , d ) d ) 45 , e ) e ) 40 | e | divide(subtract(8, add(const_2, const_3)), subtract(divide(const_1, const_2), divide(const_2, add(const_2, const_3)))) | add(const_2,const_3)|divide(const_1,const_2)|divide(const_2,#0)|subtract(n0,#0)|subtract(#1,#2)|divide(#3,#4)| | general |
a person ' s present age is one - fifth of the age of his mother . after 8 years , he will be one - half of the age of his mother . how old is the mother at present ? | "explanation : place value = local value face value = absolute value the place value of 3 in 5769354 is 3 x 100 = 300 the face value of 3 in 5769354 is nothing but 3 . = > 300 x 3 = 900 answer : option a" | a ) 900 , b ) 9000 , c ) 90 , d ) 9 , e ) 0.9 | a | multiply(multiply(3, const_1000), 3) | multiply(n0,const_1000)|multiply(n0,#0)| | general |
find the product of the place value and face value of 3 in 5769354 | "here x and y are integers . x ^ 2 = 2 y , xy = 32 . substitute ( x ^ 2 ) / 2 = y in xy = > x ^ 3 = 32 * 2 = > x ^ 3 = 64 . here x 3 is positive , x is also positive . x = 4 then y = 8 . x - y = - 4 so option d is correct" | a ) - 30 , b ) - 20 , c ) - 5 , d ) - 4 , e ) 20 | d | subtract(power(multiply(32, 2), const_0_33), divide(32, power(multiply(32, 2), const_0_33))) | multiply(n1,n2)|power(#0,const_0_33)|divide(n2,#1)|subtract(#1,#2)| | general |
if x and y are integers such that x ^ 2 = 2 y and xy = 32 , then x β y = ? | "c $ 144 given that sp = $ 102 and loss = 15 % cp = [ 100 ( sp ) ] / ( 100 - l % ) = ( 100 * 102 ) / 85 = 20 * 6 = $ 120 . to get 20 % profit , new sp = [ ( 100 + p % ) cp ] / 100 = ( 120 * 120 ) / 100 = $ 144" | a ) $ 165 , b ) $ 174 , c ) $ 144 , d ) $ 164 , e ) $ 183 | c | add(divide(102, subtract(const_1, divide(15, const_100))), multiply(divide(102, subtract(const_1, divide(15, const_100))), divide(20, const_100))) | divide(n0,const_100)|divide(n2,const_100)|subtract(const_1,#0)|divide(n1,#2)|multiply(#3,#1)|add(#3,#4)| | gain |
a shopkeeper loses 15 % , if an article is sold for $ 102 . what should be the selling price of the article to gain 20 % ? | "cp * ( 76 / 100 ) = 912 cp = 12 * 100 = > cp = 1200 answer : c" | a ) 226 , b ) 255 , c ) 1200 , d ) 2771 , e ) 332 | c | divide(912, subtract(const_1, divide(24, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain |
after decreasing 24 % in the price of an article costs rs . 912 . find the actual cost of an article ? | first selling price = 110 % - - - - - > x rupees = sold at for rs . 60 / - = 120 % - - - - - > x + 60 rupees ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 10 % - - - - - - - - > 60 100 % - - - - - - - > rs . 600 / - option ' b ' | a ) rs . 500 , b ) rs . 600 , c ) rs . 650 , d ) rs . 760 , e ) rs . 800 | b | multiply(divide(60, subtract(20, 10)), const_100) | subtract(n2,n0)|divide(n1,#0)|multiply(#1,const_100) | general |
a man sells an article at 10 % gain . had be sold at for rs . 60 / - more he could have gained 20 % what is cost price of article | "speed = ( 10 / 20 * 60 ) km / hr = ( 30 * 5 / 18 ) m / sec = 8.333333 m / sec . length of the train = 8.333333 * 6 = 50 m . answer : a" | a ) 50 , b ) 80 , c ) 120 , d ) 100 , e ) 40 | a | divide(10, subtract(divide(10, 20), 6)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics |
a train covers a distance of 10 km in 20 min . if it takes 6 sec to pass a telegraph post , then the length of the train is ? | "x = β 7 and y also = β 7 applying the function ( β 7 + β 7 ) ^ 2 - ( β 7 - β 7 ) ^ 2 = ( 2 β 7 ) ^ 2 - 0 = 4 x 7 = 28 . note : alternative approach is the entire function is represented as x ^ 2 - y ^ 2 = ( x + y ) ( x - y ) which can be simplified as ( x + y + x - y ) ( x + y - ( x - y ) ) = ( 2 x ) ( 2 y ) = 4 xy . substituting x = β 7 and y = β 7 you get the answer 28 . answer e" | a ) 0 , b ) 5 , c ) 10 , d ) 15 , e ) 28 | e | power(add(sqrt(7), sqrt(7)), 2) | sqrt(n2)|add(#0,#0)|power(#1,n0)| | general |
if x Β€ y = ( x + y ) ^ 2 - ( x - y ) ^ 2 . then β 7 Β€ β 7 = | man and days concept . . . 6 m * 80 d = m * 16 d solve it , total no of people required is 30 ; answer : c | a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | c | divide(multiply(6, 80), 16) | multiply(n0,n1)|divide(#0,n2) | physics |
6 people can do work in 80 days how much people they required to complete the work in 16 days ? | "the total subtracted is ( 9 + 8 + . . . + 1 ) = ( 9 * 10 ) / 2 = 45 on average , each number will be reduced by 45 / 10 = 4.5 therefore , the overall average will be reduced by 4.5 the answer is b ." | a ) 10 , b ) 10.5 , c ) 11 , d ) 11.5 , e ) 12 | b | divide(subtract(multiply(10, 15), multiply(add(const_4, const_1), 9)), 10) | add(const_1,const_4)|multiply(n0,n1)|multiply(n2,#0)|subtract(#1,#2)|divide(#3,n0)| | general |
the average of 10 consecutive integers is 15 . then , 9 is deducted from the first consecutive number , 8 is deducted from the second , 7 is deducted form the third , and so on until the last number which remains unchanged . what is the new average ? | "explanation : since 312 inch = 140 km so 1 inch = 140 / 312 km so 34 inch = ( 140 Γ£ β 34 ) / 312 = 15.25 km answer : d" | a ) 14.83 , b ) 14.81 , c ) 14.8 , d ) 15.25 , e ) 14.12 | d | divide(multiply(34, 140), 312) | multiply(n1,n2)|divide(#0,n0)| | physics |
on a map the distance between two mountains is 312 inches . the actual distance between the mountains is 140 km . ram is camped at a location that on the map is 34 inch from the base of the mountain . how many km is he from the base of the mountain ? | "a farmer spent 40 % money on chicken feed , so he spent 0.4 * $ 35 = $ 14 on chicken feed , thus he spent the remaining 35 - 14 = $ 21 on goat feed . now , since he bought chicken feed at a 40 % discount then the original price of it was x * 0.6 = $ 14 - - > x = $ 23,33 . therefore if the farmer had paid full price for both the chicken feed and the goat feed , then he would he have spent 23,33 + 21 = $ 44,33 . answer : d ." | a ) $ 37.80 , b ) $ 38.50 , c ) $ 39.20 , d ) $ 44.33 , e ) $ 40.60 | d | add(multiply(35, divide(40, const_100)), 35) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain |
a farmer spent $ 35 on feed for chickens and goats . he spent 40 % money on chicken feed , which he bought at a 40 % discount off the full price , and spent the rest on goat feed , which he bought at full price . if the farmer had paid full price for both the chicken feed and the goat feed , what amount would he have spent on the chicken feed and goat feed combined ? | "solution number of shares = ( 14400 / 120 ) = 120 . face value = rs . ( 100 x 120 ) = rs . 12000 . annual income = rs . ( 8 / 100 x 12000 ) = rs . 960 . answer c" | a ) rs . 500 , b ) rs . 600 , c ) rs . 960 , d ) rs . 720 , e ) none | c | multiply(multiply(100, divide(add(add(multiply(const_10, const_1000), multiply(const_4, const_1000)), multiply(const_4, 100)), add(100, multiply(100, divide(20, 100))))), divide(8, 100)) | divide(n3,n1)|divide(n2,n1)|multiply(const_10,const_1000)|multiply(const_1000,const_4)|multiply(n1,const_4)|add(#2,#3)|multiply(n1,#1)|add(#5,#4)|add(n1,#6)|divide(#7,#8)|multiply(n1,#9)|multiply(#0,#10)| | gain |
a man invested rs . 14,400 in rs . 100 shares of a company at 20 % premium . if the company declares 8 % dividend at the end of the year , then how much does he get ? | principal = ( 100 * 1500 ) / ( 12 * 20 ) = rs . 625 answer : a | a ) 625 , b ) 700 , c ) 950 , d ) 825 , e ) 630 | a | divide(divide(multiply(1500, const_100), 20), 12) | multiply(n2,const_100)|divide(#0,n1)|divide(#1,n0) | gain |
a man took loan from a bank at the rate of 12 % p . a . s . i . after 20 years he had to pay rs . 1500 interest only for the period . the principal amount borrowed by him was ? | time taken by lewis to reach tirupati = 160 / 80 = 2 hours in 2 hours , david travels 60 * 2 = 120 miles so distance at which they meet should be greater than 120 miles . only b satisfies . answer is b . | a ) 100 mlies , b ) 120 miles , c ) 90 miles , d ) 95 miles , e ) 110 miles | b | multiply(const_2, 60) | multiply(n1,const_2) | physics |
david and lewis leave chennai for tirupati simultaneously at 7 a . m in the morning driving in two cars at speeds of 60 mph and 80 mph respectively . as soon as lewis reaches tirupati he returns back to chennai along the same route and meets david on the way back . if the distance between the two cities is 160 miles , how far from chennai did david and lewis meet ? | "the number needs to be divisible by 2 ^ 2 * 3 , 2 * 7 , and 2 * 3 * 7 . the smallest such perfect square is 2 ^ 2 * 3 ^ 2 * 7 ^ 2 = 1764 the answer is d ." | a ) 900 , b ) 1089 , c ) 1600 , d ) 1764 , e ) 2025 | d | add(multiply(multiply(multiply(12, power(const_3, const_2)), 14), const_2), multiply(14, 42)) | multiply(n1,n2)|power(const_3,const_2)|multiply(n0,#1)|multiply(n1,#2)|multiply(#3,const_2)|add(#4,#0)| | geometry |
what is the smallest positive perfect square that is divisible by 12 , 14 , and 42 ? | "let the amount lent at 7 % be rs . x amount lent at 10 % is rs . ( 6000 - x ) total interest for one year on the two sums lent = 7 / 100 x + 10 / 100 ( 6000 - x ) = 600 - 3 x / 100 = > 600 - 3 / 100 x = 450 = > x = 5000 amount lent at 10 % = 1000 required ratio = 5000 : 1000 = 5 : 1 answer : a" | a ) 5 : 1 , b ) 5 : 7 , c ) 5 : 5 , d ) 5 : 9 , e ) 5 : 3 | a | divide(divide(subtract(multiply(450, const_100), multiply(6000, 7)), subtract(10, 7)), divide(subtract(multiply(450, const_100), multiply(6000, 7)), subtract(10, 7))) | multiply(n3,const_100)|multiply(n0,n1)|subtract(n2,n1)|subtract(#0,#1)|divide(#3,#2)|divide(#4,#4)| | gain |
rs . 6000 is lent out in two parts . one part is lent at 7 % p . a simple interest and the other is lent at 10 % p . a simple interest . the total interest at the end of one year was rs . 450 . find the ratio of the amounts lent at the lower rate and higher rate of interest ? | "x * y = constt . let x = y = 100 in beginning i . e . x * y = 100 * 100 = 10000 x ( 100 ) - - - becomes - - - > 1.6 x ( 120 ) i . e . 160 * new ' y ' = 10000 i . e . new ' y ' = 10000 / 160 = 62.5 i . e . y decreases from 100 to 62.5 i . e . decrease of 37.5 % answer : option d" | a ) 30 % , b ) 32 % , c ) 35 % , d ) 37.5 % , e ) 40 % | d | multiply(subtract(const_1, divide(const_100, add(const_100, 60))), const_100) | add(n0,const_100)|divide(const_100,#0)|subtract(const_1,#1)|multiply(#2,const_100)| | general |
the product of x and y is a constant . if the value of x is increased by 60 % , by what percentage must the value of y be decreased ? | if linda spent 3 / 4 of her savings on furnitute , the rest 4 / 4 - 3 / 4 = 1 / 4 on a tv but the tv cost her $ 400 . so 1 / 4 of her savings is $ 400 . so her original savings are 4 times $ 400 = $ 1600 correct answer b | a ) $ 1500 , b ) $ 1600 , c ) $ 1700 , d ) $ 1800 , e ) $ 1900 | b | divide(400, subtract(const_1, divide(3, 4))) | divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1) | general |
linda spent 3 / 4 of her savings on furniture and the rest on a tv . if the tv cost her $ 400 , what were her original savings ? | the total distance is 1 Γ 50 + 3 Γ 60 = 2301 Γ 50 + 3 Γ 60 = 230 . and the total time is 4 hours . hence , average speed = ( total distancetotal time ) = 2304 = 57.5 b | a ) 56 , b ) 57.5 , c ) 58.9 , d ) 61.4 , e ) 62 | b | divide(add(multiply(50, 1), multiply(60, 3)), add(3, 1)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0) | physics |
city a to city b , andrew drove for 1 hour at 50 mph and for 3 hours at 60 mph . what was the average speed for the whole trip ? | "explanation : we know 60 min = 1 hr total northward laxmi ' s distance = 15 kmph x 1 hr = 15 km total southward prasanna ' s distance = 45 kmph x 1 hr = 45 km total distance between prasanna and laxmi is = 15 + 45 = 60 km . answer : e" | a ) 11 , b ) 50 , c ) 28 , d ) 27 , e ) 60 | e | add(15, 45) | add(n0,n1)| | physics |
laxmi and prasanna set on a journey . laxmi moves northwards at a speed of 15 kmph and prasanna moves southward at a speed of 45 kmph . how far will be prasanna from laxmi after 60 minutes ? | in 84 minutes , car x travels 49 miles . car y gains 7 miles each hour , so it takes 7 hours to catch car x . in 7 hours , car x travels 245 miles . the answer is d . | a ) 140 , b ) 175 , c ) 210 , d ) 245 , e ) 270 | d | multiply(35, divide(multiply(divide(84, const_60), 35), subtract(42, 35))) | divide(n1,const_60)|subtract(n2,n0)|multiply(n0,#0)|divide(#2,#1)|multiply(n0,#3) | physics |
car x began traveling at an average speed of 35 miles per hour . after 84 minutes , car y began traveling at an average speed of 42 miles per hour . when both cars had traveled the same distance , both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ? | solution c ( 100 ) = 600 * 100 + 5500 = 125500 $ answer a | a ) 65500 $ , b ) 125800 $ , c ) 125900 $ , d ) 6500 $ , e ) 122500 $ | a | add(multiply(100, 600), 5500) | multiply(n0,n2)|add(n1,#0) | general |
the cost of producing x tools by a company is given by c ( x ) = 600 x + 5500 ( in $ ) a ) what is the cost of 100 tools ? | the square of an odd number is an odd number : 10 < odd < 1,000 10 < odd ^ 2 < 1,000 3 . something < odd < 31 . something ( by taking the square root ) . so , that odd number could be any odd number from 5 to 31 , inclusive : 5 , 7 , 9 , 11 , 13 , 15 , 17 , 19 , 21 , 23 , 25 , 27 , 29 , and 31 . 14 numbers . answer : c . | ['a ) 12', 'b ) 13', 'c ) 14', 'd ) 15', 'e ) 16'] | c | add(10, const_4) | add(n0,const_4) | geometry |
how many odd numbers between 10 and 1,000 are the squares of integers ? | "m = 11 s = 1.2 ds = 12.2 us = 9.8 x / 12.2 + x / 9.8 = 1 x = 5.43 d = 5.43 * 2 = 10.87 answer : e" | a ) 6.24 km , b ) 6 km , c ) 5.76 km , d ) 5.66 km , e ) 10.87 km | e | multiply(divide(multiply(add(11, 1.2), subtract(11, 1.2)), add(add(11, 1.2), subtract(11, 1.2))), const_2) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)| | physics |
a man can row 11 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and back . what is the total distance traveled by the man ? | "h . c . f = ( product of the numbers ) / ( their l . c . m ) = 38880 / 720 = 54 . answer : d" | a ) 50 , b ) 30 , c ) 125 , d ) 54 , e ) none of these | d | divide(38880, 720) | divide(n1,n0)| | physics |
if the l . c . m of two numbers is 720 and their product is 38880 , find the h . c . f of the numbers . | "a + b + c = 400 a + c = 200 b + c = 350 - - - - - - - - - - - - - - a + b + 2 c = 550 a + b + c = 400 - - - - - - - - - - - - - - - - c = 150 answer : a" | a ) a ) 150 , b ) b ) 140 , c ) c ) 130 , d ) d ) 120 , e ) e ) 110 | a | subtract(add(200, 350), 400) | add(n1,n2)|subtract(#0,n0)| | general |
a , b and c have rs . 400 between them , a and c together have rs . 200 and b and c rs . 350 . how much does c have ? | explanation : let the total sale be rs . x . then , 2.5 % . of x = 12.50 < = > ( 25 / 100 * 1 / 100 * x ) = 125 / 10 < = > x = 500 . answer : b ) 500 | a ) 333 , b ) 500 , c ) 887 , d ) 299 , e ) 132 | b | divide(12.5, divide(2.5, const_100)) | divide(n0,const_100)|divide(n1,#0) | gain |
an agent , gets a commission of 2.5 % on the sales of cloth . if on a certain day , he gets rs . 12.50 as commission , the cloth sold through him on that day is worth | an empty wooden vessel weighs 20 % of its total weight when filled with paint : vessel = 0.2 ( vessel + paint ) ; 20 v = v + p ( so the weight of completely filled vessel is 10 v ) p = 19 v ( so the weight of the paint when the vessels is completely filled is 19 v ) . the weight of a partially filled vessel is one half that of a completely filled vessel : v + p ' = 1 / 2 * 20 v ; p ' = 9 v ( so the weight of the paint when the vessels is partially filled is 9 v ) . what fraction of the vessel is filled ? so , we need to find the ratio of the weight of the paint when the vessel iscompletely filledto the weight of the paint when the vessel ispartially filled : p ' / p = 9 v / 19 v = 9 / 19 . answer : d . | a ) 3 / 5 , b ) 5 / 9 , c ) 1 / 24 , d ) 9 / 19 , e ) 2 / 5 | d | divide(subtract(divide(20, const_2), const_1), subtract(20, const_1)) | divide(n0,const_2)|subtract(n0,const_1)|subtract(#0,const_1)|divide(#2,#1) | gain |
an empty wooden vessel weighs 20 % of its total weight when filled with paint . if the weight of a partially filled vessel is one half that of a completely filled vessel , what fraction of the vessel is filled . | "900 - - - - 90 100 - - - - ? = > 10 % answer : e" | a ) 39 % , b ) 20 % , c ) 23 % , d ) 74 % , e ) 10 % | e | multiply(divide(subtract(990, 900), 900), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a cycle is bought for rs . 900 and sold for rs . 990 , find the gain percent ? | "( 0.0066 ) ( 3.6 ) / ( 0.04 ) ( 0.1 ) ( 0.006 ) = 0.0060 * 360 / 4 * ( 0.1 ) ( 0.006 ) = 0.066 * 90 / 1 * 0.006 = 66 * 90 / 6 = 11 * 90 = 990 answer : a" | a ) 990 , b ) 99.0 , c ) 9.9 , d ) 0.99 , e ) 0.099 | a | divide(multiply(0.0066, 3.6), multiply(multiply(0.04, 0.1), 0.006)) | multiply(n0,n1)|multiply(n2,n3)|multiply(n4,#1)|divide(#0,#2)| | general |
( 0.0066 ) ( 3.6 ) / ( 0.04 ) ( 0.1 ) ( 0.006 ) = | "0.09 * 0.05 = 0.0045 = 0.45 % the answer is b ." | a ) 0.15 % , b ) 0.45 % , c ) 0.8 % , d ) 1.25 % , e ) 2.0 % | b | multiply(9, divide(5, const_100)) | divide(n1,const_100)|multiply(n0,#0)| | gain |
in the manufacture of a certain product , 9 percent of the units produced are defective and 5 percent of the defective units are shipped for sale . what percent of the units produced are defective units that are shipped for sale ? | "prime numbers between 0 and 30 - 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 sum , c = 129 c / 3 = 43 answer e" | a ) 155 , b ) 129 , c ) 61 , d ) 47 , e ) 43 | e | add(divide(3, const_10), power(const_2, add(const_2, const_4))) | add(const_2,const_4)|divide(n2,const_10)|power(const_2,#0)|add(#1,#2)| | general |
let c be defined as the sum of all prime numbers between 0 and 30 . what is c / 3 | required difference = [ 3 Β½ % of rs . 9600 ] β [ 3 1 / 3 % of rs . 9600 ] = [ ( 7 / 20 - ( 10 / 3 ) ] % of rs . 9600 = 1 / 6 % of rs . 9600 = rs . [ ( 1 / 6 ) 8 ( 1 / 100 ) * 9600 ] = rs . 16 . answer is e . | a ) 11 , b ) 13 , c ) 14 , d ) 18 , e ) 16 | e | divide(multiply(subtract(add(divide(1, 2), 3), add(divide(1, 3), 3)), 9600), const_100) | divide(n1,n2)|divide(n1,n0)|add(n0,#0)|add(n0,#1)|subtract(#2,#3)|multiply(n6,#4)|divide(#5,const_100) | general |
if the sales tax reduced from 3 1 / 2 % to 3 1 / 3 % , then what difference does it make to a person who purchases an article with market price of rs . 9600 ? | for an income of rs . 8 , investment = rs . 100 . for an income of rs . 6 , investment = rs . 100 x 6 = rs . 75 . 8 market value of rs . 100 stock = rs . 75 . answer : b | a ) 33 , b ) 75 , c ) 44 , d ) 27 , e ) 91 | b | multiply(divide(const_100, 8), 6) | divide(const_100,n1)|multiply(n0,#0) | gain |
a 6 % stock yields 8 % . the market value of the stock is : | "let x be the price before the first discount . the price after the first discount is x - 25 % x ( price after first discount ) a second discount of 25 % of the discounted price after which the final price is 16 ( x - 25 % x ) - 25 % ( x - 25 % x ) = 16 solve for x x = $ 28.44 correct answer b" | a ) $ 18.44 , b ) $ 28.44 , c ) $ 48.44 , d ) $ 58.44 , e ) $ 38.44 | b | divide(multiply(multiply(const_100, const_100), 16), subtract(multiply(subtract(const_100, 25), const_100), multiply(subtract(const_100, 25), 25))) | multiply(const_100,const_100)|subtract(const_100,n0)|multiply(n2,#0)|multiply(#1,const_100)|multiply(n0,#1)|subtract(#3,#4)|divide(#2,#5)| | gain |
john bought a shirt on sale for 25 % off the original price and another 25 % off the discounted price . if the final price was $ 16 , what was the price before the first discount ? | "note : majority ( 20 % ) = difference in votes polled to win ( 60 % ) & defeated candidates ( 40 % ) 20 % = 60 % - 40 % 20 % - - - - - > 280 ( 20 Γ 14 = 280 ) 100 % - - - - - > 1400 ( 100 Γ 14 = 1400 ) a )" | a ) 1400 , b ) 1600 , c ) 1800 , d ) 2000 , e ) 2100 | a | divide(multiply(const_100, 280), subtract(60, subtract(const_100, 60))) | multiply(n1,const_100)|subtract(const_100,n0)|subtract(n0,#1)|divide(#0,#2)| | gain |
in an election between the two candidates , the candidates who gets 60 % of votes polled is winned by 280 votes majority . what is the total number of votes polled ? | "work done by 5 women in 1 day = 1 / 16 work done by 1 woman in 1 day = 1 / ( 16 Γ 5 ) work done by 16 men in 1 day = 1 / 8 work done by 1 man in 1 day = 1 / ( 8 Γ 16 ) ratio of the capacity of a man and woman = 1 / ( 8 Γ 16 ) : 1 / ( 16 Γ 5 ) = 1 / 8 : 1 / 5 = 1 / 8 : 1 / 5 = 5 : 8 option e" | a ) 1 : 3 , b ) 4 : 3 , c ) 2 : 3 , d ) 2 : 1 , e ) 5 : 8 | e | divide(divide(const_1, multiply(add(16, const_2), const_10)), divide(const_1, multiply(16, const_10))) | add(n0,const_2)|multiply(n0,const_10)|divide(const_1,#1)|multiply(#0,const_10)|divide(const_1,#3)|divide(#4,#2)| | physics |
a work can be finished in 16 days by 5 women . the same work can be finished in 8 days by sixteen men . the ratio between the capacity of a man and a woman is | given h = 60 cm and r = 5 / 2 cm total surface area = 2 Ο rh + 2 & pir ( power 2 ) = 2 Ο r ( h + r ) = [ 2 Γ 22 / 7 Γ 5 / 2 Γ ( 60 + 5 / 2 ) ] cm ( power 2 ) = [ 44 / 7 Γ 5 / 2 Γ ( ( 120 + 5 ) / 2 ) ] cm ( power 2 ) = 22 / 7 Γ 5 Γ 125 / 2 cm ( power 2 ) = ( 55 Γ 125 ) / 7 cm ( power 2 ) = 6875 / 7 cm ( power 2 ) = 982.14 cm ( power 2 ) answer is c . | ['a ) 918.14', 'b ) 981.41', 'c ) 982.14', 'd ) 928.41', 'e ) none of them'] | c | surface_cylinder(divide(5, const_2), 60) | divide(n1,const_2)|surface_cylinder(#0,n0) | geometry |
the height of a cylinder is 60 cm and the diameter of its base is 5 cm . the total surface area of the cylinder is | "suppose pipe a alone take x hours to fill the tank then pipe b and c will take x / 2 and x / 4 hours respectively to fill the tank . 1 / x + 2 / x + 4 / x = 1 / 7 7 / x = 1 / 7 x = 49 hours answer is d" | a ) 25 hr , b ) 35 hr , c ) 40 hr , d ) 49 hr , e ) 50 hr | d | multiply(add(add(multiply(const_2, const_2), const_2), const_1), 7) | multiply(const_2,const_2)|add(#0,const_2)|add(#1,const_1)|multiply(n1,#2)| | physics |
a tank is filled by 3 pipes a , b , c in 7 hours . pipe c is twice as fast as b and b is twice as fast as a . how much will pipe a alone take to fill the tank ? | "let the two positive numbers be 5 x and 8 x respectively . 8 x - 5 x = 27 3 x = 27 = > x = 9 = > smaller number = 5 x = 45 . answer : c" | a ) 25 , b ) 66 , c ) 45 , d ) 88 , e ) 44 | c | divide(multiply(27, 5), const_4) | multiply(n0,n2)|divide(#0,const_4)| | other |
there are two positive numbers in the ratio 5 : 8 . if the larger number exceeds the smaller by 27 , then find the smaller number ? | amount earned using her cell phone = 70 * 22 = 1540 earned for remaining surveys = 30 * 20 = 600 total earning = 2140 answer : a | a ) 2140 , b ) 1140 , c ) 550 , d ) 650 , e ) 750 | a | add(multiply(20, 100), multiply(70, multiply(20, divide(10, const_100)))) | divide(n2,const_100)|multiply(n0,n1)|multiply(n0,#0)|multiply(n3,#2)|add(#1,#3) | gain |
a worker is paid a regular rate of rs . 20 for completing a survey . the worker completes 100 surveys per week . for any survey involving the use of her cellphone , she is paid at a rate of that is 10 % higher than her regular rate . if she completed 70 surveys involving the use of her cellphone , how much did she get that week ? | "0.1 x = 0.08 ( 2000 - x ) + 38 0.18 x = 198 x = 1100 then the amount invested at 8 % is $ 2000 - $ 1100 = $ 900 the answer is c ." | a ) $ 700 , b ) $ 800 , c ) $ 900 , d ) $ 1100 , e ) $ 1200 | c | subtract(multiply(multiply(const_100, 10), const_2), divide(add(multiply(multiply(10, 8), const_2), 38), add(divide(10, const_100), divide(8, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|multiply(n0,const_100)|multiply(n0,n1)|add(#0,#1)|multiply(#2,const_2)|multiply(#3,const_2)|add(n4,#6)|divide(#7,#4)|subtract(#5,#8)| | general |
if x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year , the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $ 38 . if $ 2,000 is the total amount invested , how much is invested at 8 percent ? | "in 4 miles , initial 1 / 5 mile charge is $ 2 rest of the distance = 4 - ( 1 / 5 ) = 19 / 5 rest of the distance charge = 19 ( 0.4 ) = $ 7.6 ( as the charge is 0.4 for every 1 / 5 mile ) = > total charge for 4 miles = 2 + 7.6 = 9.6 answer is d" | a ) $ 4.60 , b ) $ 9.80 , c ) $ 19.60 , d ) $ 9.60 , e ) $ 29.60 | d | add(2.00, multiply(subtract(divide(2.00, divide(1, 5)), 1), 0.40)) | divide(n1,n2)|divide(n0,#0)|subtract(#1,n1)|multiply(n3,#2)|add(n0,#3)| | general |
if taxi fares were $ 2.00 for the first 1 / 5 mile and $ 0.40 for each 1 / 5 mile there after , then the taxi fare for a 4 - mile ride was | "the ratio of the times taken is 2 : 1 . the ratio of the speed of the boat in still water to the speed of the stream = ( 2 + 1 ) / ( 2 - 1 ) = 3 / 1 = 3 : 1 speed of the stream = 42 / 3 = 14 kmph . answer : c" | a ) 16 kmph , b ) 18 kmph , c ) 14 kmph , d ) 79 kmph , e ) 27 kmph | c | subtract(42, divide(multiply(42, const_2), const_3)) | multiply(n0,const_2)|divide(#0,const_3)|subtract(n0,#1)| | physics |
the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is 42 kmph , find the speed of the stream ? | "sp = 300 cp = ( sp ) * [ 100 / ( 100 + p ) ] = 300 * [ 100 / ( 100 + 25 ) ] = 300 * [ 100 / 125 ] = rs . 240 answer : c" | a ) s . 486 , b ) s . 455 , c ) s . 240 , d ) s . 480 , e ) s . 489 | c | divide(multiply(300, const_100), add(const_100, 25)) | add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)| | gain |
by selling an article at rs . 300 , a profit of 25 % is made . find its cost price ? | "explanation : ( 25000 Γ ( 1 + 12100 ) 3 ) = > 25000 Γ 2825 Γ 2825 Γ 2825 = > 35123.20 so compound interest will be 35123.20 - 25000 = rs 10123.20 answer : a" | a ) rs 10123.20 , b ) rs 10123.30 , c ) rs 10123.40 , d ) rs 10123.50 , e ) none of these | a | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | divide(n2,const_100)|multiply(const_100,const_4)|add(#0,const_1)|multiply(#1,const_100)|power(#2,n1)|multiply(#3,#4)|subtract(#5,#3)| | gain |
what will be the compound interest on rs . 25000 a Ε er 3 years at the rate of 12 % per annum | number of runs scored more to increase the ratio by 1 is 26 - 14 = 12 to raise the average by one ( from 14 to 15 ) , he scored 12 more than the existing average . therefore , to raise the average by five ( from 14 to 19 ) , he should score 12 x 5 = 60 more than the existing average . thus he should score 14 + 60 = 74 . answer d | a ) 12 , b ) 18 , c ) 25 , d ) 74 , e ) 88 | d | subtract(multiply(19, add(subtract(26, 15), const_1)), multiply(14, subtract(26, 15))) | subtract(n0,n2)|add(#0,const_1)|multiply(n1,#0)|multiply(n3,#1)|subtract(#3,#2) | general |
a batsman scores 26 runs and increases his average from 14 to 15 . find the runs to be made if he wants top increasing the average to 19 in the same match ? | "explanation : ( a + b ) β s 10 days work = 10 [ 1 / 30 + 1 / 30 ] = 10 [ 1 + 1 / 30 ] = 2 / 3 a complete remaining work in 2 / 3 * 30 = 20 total work = 10 + 20 = 30 days answer : option a" | a ) 30 days , b ) 35 days , c ) 40 days , d ) 45 days , e ) 50 days | a | divide(subtract(10, add(divide(10, 30), divide(10, 30))), divide(10, 30)) | divide(n2,n1)|divide(n2,n0)|add(#0,#1)|subtract(n2,#2)|divide(#3,#0)| | physics |
a and b can do a piece of work in 30 days and 30 days respectively . they work together for 10 days and b leaves . in how many days the whole work is completed ? | there are 520 players , only 1 person wins , 519 players lose . in order to lose , you must have lost a game . 519 games . ans - b | a ) 511 , b ) 519 , c ) 256 , d ) 255 , e ) 1023 | b | add(add(add(add(add(add(add(divide(divide(divide(520, const_2), const_2), const_2), add(divide(520, const_2), divide(divide(520, const_2), const_2))), divide(divide(divide(divide(520, const_2), const_2), const_2), const_2)), divide(divide(divide(divide(divide(520, const_2), const_2), const_2), const_2), const_2)), divide(divide(divide(divide(divide(divide(520, const_2), const_2), const_2), const_2), const_2), const_2)), divide(divide(divide(divide(divide(divide(divide(520, const_2), const_2), const_2), const_2), const_2), const_2), const_2)), divide(divide(divide(divide(divide(divide(divide(divide(520, const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2)), divide(divide(divide(divide(divide(divide(divide(divide(divide(520, const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2)) | divide(n0,const_2)|divide(#0,const_2)|add(#0,#1)|divide(#1,const_2)|add(#2,#3)|divide(#3,const_2)|add(#4,#5)|divide(#5,const_2)|add(#6,#7)|divide(#7,const_2)|add(#8,#9)|divide(#9,const_2)|add(#10,#11)|divide(#11,const_2)|add(#12,#13)|divide(#13,const_2)|add(#14,#15) | general |
a total of 520 players participated in a single tennis knock out tournament . what is the total number of matches played in the tournament ? ( knockout means if a player loses , he is out of the tournament ) . no match ends in a tie . | ( 7 w + 6 ) / 6 + ( 9 w + 8 ) / 2 = 22 or , [ 7 w + 6 + 3 ( 9 w + 8 ) ] / 6 = 22 or , 7 w + 6 + 27 w + 24 = 132 or , 34 w + 30 = 132 or , 34 w = 132 - 30 or , 34 w = 102 or , w = 102 / 34 therefore , w = 3 answer : c | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(subtract(multiply(multiply(2, 6), 22), add(multiply(8, 6), multiply(2, 6))), add(multiply(9, 6), multiply(7, const_2))) | multiply(n1,n5)|multiply(n1,n4)|multiply(n1,n3)|multiply(n0,const_2)|add(#1,#0)|add(#2,#3)|multiply(n6,#0)|subtract(#6,#4)|divide(#7,#5) | general |
solving a linear equation with several occurrences of the variable , solve for w . simplify answer as much as possible . ( 7 w + 6 ) / 6 + ( 9 w + 8 ) / 2 = 22 | "answer = b = 2 f ( x ) = 3 x β 5 2 * [ f ( x ) ] β 7 = f ( 3 x β 6 ) 2 ( 3 x - 5 ) - 7 = 3 ( 3 x - 6 ) - 5 6 x - 17 = 9 x - 23 x = 2" | a ) 0 , b ) 2 , c ) 6 , d ) 7 , e ) 13 | b | divide(subtract(add(multiply(2, 5), 7), add(multiply(3, 3), 5)), subtract(multiply(2, 3), multiply(3, const_1))) | multiply(n1,n2)|multiply(n4,n0)|multiply(n0,n2)|multiply(n0,const_1)|add(n3,#0)|add(n1,#1)|subtract(#2,#3)|subtract(#4,#5)|divide(#7,#6)| | general |
given f ( x ) = 3 x β 5 , for what value of x does 2 * [ f ( x ) ] β 7 = f ( 3 x β 6 ) | "sol . relative speed = ( 45 + 30 ) km / hr = ( 75 x 5 / 18 ) m / sec = ( 125 / 6 ) m / sec . distance covered = ( 750 + 750 ) m = 1500 m . required time = ( 1500 x 6 / 125 ) sec = 72 sec . answer d" | a ) 12 sec , b ) 24 sec , c ) 48 sec , d ) 72 sec , e ) none | d | multiply(multiply(750, inverse(multiply(add(45, 30), const_0_2778))), const_2) | add(n1,n2)|multiply(#0,const_0_2778)|inverse(#1)|multiply(n0,#2)|multiply(#3,const_2)| | physics |
two good train each 750 m long , are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one . | let the length and the width be 4 x and 3 x respectively . area = ( 4 x ) ( 3 x ) = 4800 12 x ^ 2 = 4800 x ^ 2 = 400 x = 20 the ratio of the width and the area is 3 x : 12 x ^ 2 = 1 : 4 x = 1 : 80 the answer is c . | ['a ) 1 : 72', 'b ) 1 : 76', 'c ) 1 : 80', 'd ) 1 : 84', 'e ) 1 : 88'] | c | divide(divide(sqrt(multiply(const_3, 4800)), const_2), 4800) | multiply(n2,const_3)|sqrt(#0)|divide(#1,const_2)|divide(#2,n2) | geometry |
the ratio of the length and the width of a rectangle is 4 : 3 and the area of the rectangle is 4800 sq cm . what is the ratio of the width and the area of the rectangle ? | "x - - - - - - - 7 ( x + 10 ) - - - - 6 x * 7 = ( x + 10 ) 6 x = 60 \ answer : c" | a ) 22 , b ) 20 , c ) 60 , d ) 71 , e ) 11 | c | divide(multiply(multiply(3, const_2), 10), subtract(7, multiply(3, const_2))) | multiply(n1,const_2)|multiply(n2,#0)|subtract(n0,#0)|divide(#1,#2)| | physics |
a work which could be finished in 7 days was finished 3 days earlier after 10 more men joined . the number of men employed was ? | x can have at most 3 prime factors , namely the prime factors 2 and 3 , plus one other . if x had more than this number of prime factors , then ( 6 ) ( x ^ 2 ) would have more than 3 prime factors . the answer is c . | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | multiply(3, const_1) | multiply(n2,const_1) | general |