Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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3 years ago , the ratio of ages of p and q was 4 : 3 . 6 years from now , the ratio of their ages will be 7 : 6 . what is p ' s age at present ? | 3 ( p - 3 ) / 4 = ( q - 3 ) . then q = 3 ( p - 3 ) / 4 + 3 6 ( p + 6 ) / 7 = q + 6 6 ( p + 6 ) / 7 = 3 ( p - 3 ) / 4 + 9 24 p + 144 = 21 p + 189 3 p = 45 p = 15 the answer is c . | a ) 9 , b ) 12 , c ) 15 , d ) 18 , e ) 21 | c | add(divide(subtract(multiply(add(3, 6), 7), multiply(add(3, 6), 6)), subtract(6, multiply(7, divide(3, 4)))), 3) | add(n0,n3)|divide(n0,n1)|multiply(n4,#0)|multiply(n3,#0)|multiply(n4,#1)|subtract(#2,#3)|subtract(n3,#4)|divide(#5,#6)|add(n0,#7) | other | C |
45 workers work 8 hours to dig a hole 30 meters deep . how many extra workers should be hired to dig another hole 45 meters deep by working for 6 hours ? | "45 workers * 8 hours / 30 meters = x * 6 / 45 x = 90 total workers 90 - 45 = 45 new workers the answer is c ." | a ) 25 , b ) 35 , c ) 45 , d ) 55 , e ) 65 | c | subtract(multiply(multiply(45, divide(8, 6)), divide(45, 30)), 45) | divide(n3,n2)|divide(n1,n4)|multiply(n0,#1)|multiply(#0,#2)|subtract(#3,n0)| | physics | C |
some of 15 % - intensity red paint is replaced with 25 % solution of red paint such that the new paint intensity is 30 % . what fraction of the original paint was replaced ? | "let total paint = 1 let amount replaced = x 15 ( 1 - x ) + 25 x = 30 x = 3 / 2 answer : d" | a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 3 / 2 , e ) 3 / 5 | d | divide(subtract(divide(30, const_100), divide(15, const_100)), subtract(divide(25, const_100), divide(15, const_100))) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(#2,#1)|divide(#3,#4)| | gain | D |
at a garage sale , all of the items were sold at different prices . if the price of a radio sold at the garage sale was both the 18 th highest price and the 25 th lowest price among the prices of the items sold , how many items were sold at the garage sale ? | "there were 17 items sold at a higher price than the radio and 24 items sold at a lower price than the radio . including the radio , there were 17 + 24 + 1 = 42 items sold . the answer is b ." | a ) 41 , b ) 42 , c ) 43 , d ) 44 , e ) 45 | b | subtract(add(25, 18), const_1) | add(n0,n1)|subtract(#0,const_1)| | other | B |
what is the sum of the numbers between 1 and 20 , inclusive ? | "sol . 1 + 2 + 3 . . . 17 + 18 + 19 + 20 answer . e" | a ) 310 , b ) 200 , c ) 215 , d ) 285 , e ) 210 | e | divide(multiply(1, 20), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general | E |
two equally sized jugs full of water are each emptied into two separate unequally sized empty jugs , x and y . now , jug x is 1 / 4 full , while jug y is 2 / 3 full . if water is poured from jug x into jug y until jug y is filled , what fraction of jug x then contains water ? | "suppose the water in each jug is l liters cx x ( 1 / 4 ) = l cx = 4 l liters cx is capacity of x cy x ( 2 / 3 ) = l cy = 3 l / 2 liters cy is capacity of y now , y is 3 l / 2 - l empty = l / 2 empty so , we can put only l / 2 water in jug y from jug x jug x ' s remaining water = l - l / 2 = l / 2 fraction of x which contains water = water / cx = ( l / 2 ) / 4 l = 1 / 8 answer will be d" | a ) 0 , b ) 1 / 15 , c ) 2 / 15 , d ) 1 / 8 , e ) 2 / 10 | d | multiply(divide(1, 4), divide(subtract(1, divide(2, 3)), divide(2, 3))) | divide(n0,n1)|divide(n2,n3)|subtract(n0,#1)|divide(#2,#1)|multiply(#0,#3)| | general | D |
how many cubes of 50 cm edge can be put in a cubical box of 1 m edge . | "number of cubes = 100 â ˆ — 100 â ˆ — 100 / 50 * 50 * 50 = 8 number of cubes = 100 â ˆ — 100 â ˆ — 10010 â ˆ — 10 â ˆ — 10 = 1000 note : 1 m = 100 cm answer : c" | a ) 17 cm , b ) 10 cm , c ) 8 cm , d ) 6 cm , e ) 7 cm | c | divide(volume_cube(1), volume_cube(divide(50, const_100))) | divide(n0,const_100)|volume_cube(n1)|volume_cube(#0)|divide(#1,#2)| | physics | C |
a certain telephone company offers two plans , a and b . under plan a , the company charges a total of $ 0.60 for the first 8 minutes of each call and $ 0.06 per minute thereafter . under plan b , the company charges $ 0.08 per minute of each call . what is the duration of a call , in minutes , for which the company charges the same amount under plan a and under plan b ? | let the duration , in minutes , for which the company charges the same under plan a and plan b be t minutes . then under plan a the cost would be $ 0.6 + 0.06 ( t - 8 ) and under plan b the cost would be $ 0.08 t . we want these amount to be equal : 0.6 + 0.06 ( t - 8 ) = 0.08 t - - > 60 + 6 ( t - 8 ) = 8 t - - > t = 6 . answer : c . | a ) 2 , b ) 9 , c ) 6 , d ) 21 , e ) 30 | c | divide(subtract(0.6, multiply(0.06, 8)), subtract(0.08, 0.06)) | multiply(n1,n2)|subtract(n3,n2)|subtract(n0,#0)|divide(#2,#1) | other | C |
a present value of a machine is $ 1200 . its value depletion rate is 10 % per annum then find the machine value after 2 years ? | "p = $ 1200 r = 10 % t = 2 years machine value after 2 years = p [ ( 1 - r / 100 ) ^ t ] = 1200 * 9 / 10 * 9 / 10 = $ 972 answer is a" | a ) $ 972 , b ) $ 810 , c ) $ 915 , d ) $ 715 , e ) $ 795 | a | multiply(1200, power(subtract(const_1, divide(10, const_100)), 2)) | divide(n1,const_100)|subtract(const_1,#0)|power(#1,n2)|multiply(n0,#2)| | gain | A |
the current in a river is 4 mph . a boat can travel 18 mph in still water . how far up the river can the boat travel if the round trip is to take 10 hours ? | "upstream speed = 18 - 4 = 14 mph downstream speed = 18 + 4 = 22 mph d / 14 + d / 22 = 10 hours solving for d we get d = 85.5 answer : b" | a ) 69 miles , b ) 85.5 miles , c ) 96 miles , d ) 100 miles , e ) 112 miles | b | divide(multiply(multiply(subtract(18, 4), add(18, 4)), 10), add(add(18, 4), subtract(18, 4))) | add(n0,n1)|subtract(n1,n0)|add(#0,#1)|multiply(#0,#1)|multiply(n2,#3)|divide(#4,#2)| | physics | B |
a tank is 25 m long 12 m wide and 6 m deep . the cost of plastering its walls and bottom at 45 paise per sq m is | "explanation : area to be plastered = [ 2 ( l + b ) ã — h ] + ( l ã — b ) = [ 2 ( 25 + 12 ) ã — 6 ] + ( 25 ã — 12 ) = 744 sq m cost of plastering = 744 ã — ( 45 / 100 ) = rs . 334.80 answer : b" | a ) rs . 234.80 , b ) rs . 334.80 , c ) rs . 434.80 , d ) rs . 534.80 , e ) none of these | b | multiply(divide(45, const_100), add(multiply(25, 12), add(multiply(const_2, multiply(25, 6)), multiply(multiply(12, 6), const_2)))) | divide(n3,const_100)|multiply(n0,n2)|multiply(n1,n2)|multiply(n0,n1)|multiply(#1,const_2)|multiply(#2,const_2)|add(#4,#5)|add(#6,#3)|multiply(#7,#0)| | physics | B |
the area of one square is x ^ 2 + 10 x + 25 and the area of another square is 4 x ^ 2 − 12 x + 9 . if the sum of the perimeters of both squares is 32 , what is the value of x ? | "spotting the pattern of equations both are in form of ( x + c ) ^ 2 so a 1 = ( x + 5 ) ^ 2 & a 2 = ( 2 x - 3 ) ^ 2 l 1 = x + 5 & l 2 = 2 x - 3 p 1 = 4 ( x + 5 ) & p 2 = 4 ( 2 x - 3 ) p 1 + p 2 = 32 4 ( x + 5 ) + 4 ( 2 x - 3 ) = 32 . . . . . . . . . . . . . . > x = 2 answer : b" | a ) 0 , b ) 2 , c ) 2.5 , d ) 4.67 , e ) 10 | b | divide(subtract(32, subtract(multiply(4, divide(10, 2)), 10)), 10) | divide(n1,n0)|multiply(#0,n3)|subtract(#1,n1)|subtract(n7,#2)|divide(#3,n1)| | general | B |
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 | B |
the price of a jacket is reduced by 10 % . during a special sale the price of the jacket is reduced another 30 % . 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 10 % , therefore bringing down the price to $ 90 . 3 ) again it is further discounted by 30 % , therefore bringing down the price to $ 63 . 4 ) now 67.5 has to be added byx % in order to equal the original price . 63 + ( x % ) 63 = 100 . solving this eq for x , we get x = 58.7 ans is b" | a ) 32.5 , b ) 58.7 , c ) 48 , d ) 65 , e ) 67.5 | b | multiply(const_100, divide(subtract(const_100, subtract(subtract(const_100, 10), multiply(subtract(const_100, 10), divide(30, const_100)))), subtract(subtract(const_100, 10), multiply(subtract(const_100, 10), divide(30, 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 | B |
the average of 65 numbers is 40 . if two numbers , 83 and 30 are discarded , then the average of the remaining numbers is nearly | explanation : total sum of 63 numbers = ( 65 * 40 ) - ( 83 + 30 ) = 2600 - 113 = 2487 average = 2487 / 63 = 39.47 answer : e | a ) 28.32 , b ) 39 , c ) 38.78 , d ) 29 , e ) 39.47 | e | divide(subtract(multiply(65, 40), add(83, 30)), subtract(65, const_2)) | add(n2,n3)|multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,#0)|divide(#3,#2) | general | E |
in how many ways can 5 people be arranged in a circle ? | "( 5 − 1 ) ! = 4 ! = 24 ways ans - c" | a ) 2 ! , b ) 5 ! , c ) 4 ! , d ) 3 ! , e ) 6 ! | c | subtract(divide(divide(5, const_1), const_3), const_3) | divide(n0,const_1)|divide(#0,const_3)|subtract(#1,const_3)| | probability | C |
how many multiples of 5 are there between 1 and 100 ( both are inclusive ) ? | "the answer is ( 100 - 1 ) / 5 + 1 = 20.8 answer is b" | a ) 1.2 , b ) 20.8 , c ) 2.3 , d ) 6.7 , e ) 8.2 | b | add(divide(subtract(100, 1), 5), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | other | B |
beginning in town a , biker bob rides his bike 8 miles west , 5 miles north , 4 miles east , and then 15 miles north , to town b . how far apart are town a and town b ? ( ignore the curvature of the earth . ) | "using pythagoras we have one side i , e total distance traveled in north direction = 15 + 5 = 20 m other being the base ie distance traveled west - distance traveled eat = 8 - 4 = 4 m now this third side or the distance between town a and town b = 20 ^ 2 + 4 ^ 2 = sq root 416 = 20.4 miles answer : b" | a ) 25 miles , b ) 20.4 miles , c ) 30 miles , d ) 32 miles , e ) 33 miles | b | sqrt(add(power(add(5, 15), const_2), power(4, const_2))) | add(n1,n3)|power(n2,const_2)|power(#0,const_2)|add(#2,#1)|sqrt(#3)| | physics | B |
a candidate got 40 % of the votes polled and he lost to his rival by 2000 votes . how many votes were cast ? | "40 % - - - - - - - - - - - l 60 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 20 % - - - - - - - - - - 2000 100 % - - - - - - - - - ? = > 10000 answer : a" | a ) 10000 , b ) 2028 , c ) 2775 , d ) 5496 , e ) 6851 | a | divide(2000, subtract(subtract(const_1, divide(40, const_100)), divide(40, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n1,#2)| | gain | A |
a person distributed 15 % of his income to his 3 children each . he deposited 30 % of his income to his wife ' s account . he donated 10 % of remaining amount to an orphan house . finally he has $ 40000 . find his total income ? | 3 children got = 3 * 15 % = 45 % wife got = 30 % orphan house = 10 % total = 45 + 30 + 10 = 85 % remaining = 100 - 85 = 15 % 15 % = 40000 100 % = 40000 * 100 / 15 = $ 266666.7 answer is a | a ) a ) 266666.7 , b ) b ) 562000 , c ) c ) 800000 , d ) d ) 500000 , e ) e ) 652000 | a | multiply(divide(40000, subtract(const_100, add(add(multiply(15, 3), 30), 10))), const_100) | multiply(n0,n1)|add(n2,#0)|add(n3,#1)|subtract(const_100,#2)|divide(n4,#3)|multiply(#4,const_100) | gain | A |
how many liters of water must be evaporated from 50 liters of a 3 percent sugar solution to get a 5 percent sugar solution ? | "3 % sugar solution means that currently there are 1.5 liters of sugar and this amount must be 5 % of a reduced final amount ( 50 - x ) 1.5 = 0.05 * ( 50 - x ) 0.05 x = 1 x = 20 answer e" | a ) 2 , b ) 4 , c ) 6 , d ) 10 , e ) 20 | e | subtract(50, multiply(divide(50, const_100), 5)) | divide(n0,const_100)|multiply(n2,#0)|subtract(n0,#1)| | gain | E |
a clock store sold a certain clock to a collector for 30 percent more than the store had originally paid for the clock . when the collector tried to resell the clock to the store , the store bought it back at 60 percent of what the collector had paid . the shop then sold the clock again at a profit of 90 percent on its buy - back price . if the difference between the clock ' s original cost to the shop and the clock ' s buy - back price was $ 150 , for how much did the shop sell the clock the second time ? | "now , in the question above , lets say the original cost of the clock to store was c $ and then it sold the same to the collector at 30 % profit . this means the clocks ' selling price was c ( 1.3 ) and this becomes cost price for the collector . now , when the collector tries to sell the same clock to the store , the store buys it for 60 % the price at which the collector bought it . thus , you get = 1.3 * 0.6 * c = 0.78 c furthermore , the store sells the clock for the second time for 90 % profit and thus the selling price of the clock becomes = cost price of the clock for the store at buy - back * 1.9 = 1.9 * 0.78 c finally given that c - 0.78 c = 150 - - - - > c = 681.82 $ thus , the cost of the clock the second time around = 1.9 * 0.78 c = 1.9 * 0.78 * 681.82 = 1010.46 $ . hence a is the correct answer ." | a ) $ 1010.46 , b ) $ 505.23 , c ) $ 2020.92 , d ) $ 800.46 , e ) $ 1515.69 | a | divide(multiply(150, divide(multiply(add(150, 90), divide(add(150, 30), const_2)), 150)), subtract(divide(add(150, 30), const_2), 30)) | add(n2,n3)|add(n0,n3)|divide(#1,const_2)|multiply(#0,#2)|subtract(#2,n0)|divide(#3,n3)|multiply(n3,#5)|divide(#6,#4)| | general | A |
in a kilometer race , a beats b by 20 meters or 10 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 10 ) / 20 = 500 sec . time taken by a = 500 - 10 = 490 sec . answer : e" | a ) 290 sec , b ) 190 sec , c ) 199 sec , d ) 192 sec , e ) 490 sec | e | subtract(divide(multiply(const_1, const_1000), divide(20, 10)), 10) | divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)| | physics | E |
two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.50 km and 1.0 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | ": relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.50 + 1.0 = 2.5 km = 2500 m . required time = 2500 * 3 / 125 = 60 sec . answer : d" | a ) 48 , b ) 9 , c ) 7 , d ) 60 , e ) 15 | d | subtract(divide(multiply(1.50, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.0, const_1000), divide(multiply(90, const_1000), const_3600))) | multiply(n2,const_1000)|multiply(n0,const_1000)|multiply(n3,const_1000)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#3,const_3600)|divide(#0,#4)|divide(#2,#5)|subtract(#6,#7)| | physics | D |
of 25 applicants for a job , 14 had at least 4 years ' experience , 18 had degrees , and 3 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | "set a : people with more than 4 years exp set b : people with degree aub = total - ( less than 4 exp and no degree ) aub = 25 - 3 = 22 aub = a + b - aib aib = 18 + 14 - 22 = 10 answer b" | a ) 14 , b ) 10 , c ) 9 , d ) 7 , e ) 5 | b | add(subtract(add(14, 18), subtract(25, 3)), subtract(18, 14)) | add(n1,n3)|subtract(n0,n4)|subtract(n3,n1)|subtract(#0,#1)|add(#3,#2)| | general | B |
in a kilometer race , a beats b by 54 meters or 12 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 12 ) / 54 = 222 sec . time taken by a = 224 - 12 = 212 sec . answer : a" | a ) 212 , b ) 190 , c ) 277 , d ) 181 , e ) 122 | a | subtract(divide(multiply(const_1, const_1000), divide(54, 12)), 12) | divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)| | physics | A |
a train 280 m long , running with a speed of 72 km / hr will pass a tree in ? | "speed = 72 * 5 / 18 = 20 m / sec time taken = 280 * 1 / 20 = 14 sec answer : d" | a ) 17 sec , b ) 16 sec , c ) 18 sec , d ) 14 sec , e ) 12 sec | d | multiply(divide(280, multiply(72, const_1000)), const_3600) | multiply(n1,const_1000)|divide(n0,#0)|multiply(#1,const_3600)| | physics | D |
if y is the smallest positive integer such that 1260 multiplied by y is the square of an integer , then y must be | "1260 = 2 ^ 2 * 3 ^ 2 * 5 * 7 to be perfect square , we need to multiply by at least 5 * 7 = 35 . the answer is b ." | a ) 25 , b ) 35 , c ) 60 , d ) 75 , e ) 140 | b | multiply(add(const_2, const_3), const_2) | add(const_2,const_3)|multiply(#0,const_2)| | geometry | B |
what is the measure of the radius of the circle that circumscribes a triangle whose sides measure 3 , 4 , and 5 ? | "some of pyhtagron triplets we need to keep it in mind . like { ( 2 , 3,5 ) , ( 5 , 12,13 ) , ( 7 , 24,25 ) , ( 11 , 60,61 ) . so now we know the triangle is an right angle triangle . the circle circumscribes the triangle . the circum raduis of the circle that circumscribes the right angle triangle = hypotanse / 2 = 5 / 2 = 2.5 ans . e" | a ) 3.5 , b ) 3 , c ) 5 , d ) 4 , e ) 2.5 | e | divide(triangle_area_three_edges(3, 4, 5), divide(triangle_perimeter(3, 4, 5), const_2)) | triangle_area_three_edges(n0,n1,n2)|triangle_perimeter(n0,n1,n2)|divide(#1,const_2)|divide(#0,#2)| | geometry | E |
pipe a fills a swimming pool in 4 hours . pipe b empties the pool in 6 hours . if pipe a was opened at 5 : 00 am and pipe b at 6 : 00 am , at what time will the pool be full ? | "pipe a fills the pool in 4 hrs . 1 hour ' s work : 1 / 4 pipe b empties the pool in 6 hrs . 1 hour ' s work : 1 / 6 together if they work , 1 hour ' s work = 1 / 4 - 1 / 6 = 1 / 12 given : pipe a started at 5 : 00 a . m and pipe b at 6 : 00 a . m pool filled after 1 hour by pipe a : 1 / 4 or 3 / 12 after 6 : 00 a . m pool filled after 1 hour with both the pipes on : 1 / 12 pool filled after 9 hours with both pipes on : 9 / 12 pool filled in 1 hour + pool filled in 9 hours = 3 / 12 + 9 / 12 = 1 therefore , it takes 10 hrs to fill the pool as pipe a started at 5 : 00 a . m , pool is full at 15 : 00 hrs answer : c" | a ) 13 : 00 , b ) 14 : 00 , c ) 15 : 00 , d ) 17 : 00 , e ) 16 : 00 | c | multiply(divide(const_3, 4), divide(const_1, subtract(divide(const_1, 4), divide(const_1, 6)))) | divide(const_3,n0)|divide(const_1,n0)|divide(const_1,n1)|subtract(#1,#2)|divide(const_1,#3)|multiply(#0,#4)| | physics | C |
n 10 years , a will be twice as old as b was 10 years ago . if a is now 9 years older than b the present age of b is | a + 10 = 2 ( b - 10 ) . . . . . . . . ( 1 ) a = b + 9 . . . . . . . . . ( 2 ) from equations . 1 & 2 we get b = 39 a will be 39 + 9 = 48 years old . answer : e | a ) 22 , b ) 897 , c ) 268 , d ) 20 , e ) 48 | e | add(9, add(add(multiply(const_2, 10), 10), 9)) | multiply(n0,const_2)|add(n0,#0)|add(n2,#1)|add(n2,#2) | general | E |
what is the median of a set of consecutive integers if the sum of nth number from the beginning and nth number from the end is 50 ? | "surprisingly no one answered this easy one . property of a set of consecutive integerz . mean = median = ( first element + last element ) / 2 = ( second element + last but one element ) / 2 = ( third element + third last element ) / 2 etc . etc . so mean = median = 50 / 2 = 25 answer is b" | a ) 10 , b ) 25 , c ) 50 , d ) 75 , e ) 100 | b | divide(50, const_2) | divide(n0,const_2)| | general | B |
in a class , 30 students pass in english and 20 students in math , while some students among these pass in both . how many more students do only english as compared to those doing only maths ? | let us consider tht x student are tohse who passed in both english and maths . . . so first we remove x student from both of them therefore , english = 30 - x maths = 20 - x now , number of students more in english = ( 30 - x ) - ( 20 - x ) = 30 - x - 20 + x = 10 answer : a | a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) indeterminate | a | subtract(30, 20) | subtract(n0,n1) | other | A |
an engineer designed a ball so that when it was dropped , it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a 16 - meter platform and caught it after it had traveled 32.5 meters . how many times did the ball bounce ? | "ans : 2 division of total diatance travelled will be 16 + 16 + 0.5 a" | a ) 2 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | a | divide(divide(16, const_2), const_2) | divide(n0,const_2)|divide(#0,const_2)| | general | A |
the ratio 3 : 5 expressed as a percent equals | "solution 3 : 5 = 3 / 5 = ( 3 / 5 x 100 ) % . = 60 % . answer a" | a ) 60 % , b ) 40 % , c ) 80 % , d ) 125 % , e ) none | a | multiply(divide(3, 5), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | A |
what is the sum of all the prime numbers from 20 to 40 ? | "explanation : 23 + 29 + 31 + 37 = 120 answer : d" | a ) 279 , b ) 100 , c ) 278 , d ) 120 , e ) 191 | d | multiply(add(divide(subtract(subtract(40, 20), const_2), const_2), 20), divide(add(subtract(40, 20), const_2), const_2)) | subtract(n1,n0)|add(#0,const_2)|subtract(#0,const_2)|divide(#2,const_2)|divide(#1,const_2)|add(n0,#3)|multiply(#5,#4)| | general | D |
the average of 10 numbers is 40.2 . later it is found that two numbers have been wrongly copied . the first is 16 greater than the actual number and the second number added is 13 instead of 31 . find the correct average . | "sum of 10 numbers = 402 corrected sum of 10 numbers = 402 – 13 + 31 – 16 = 404 hence , new average = 404 ⁄ 10 = 40.4 answer b" | a ) 40.2 , b ) 40.4 , c ) 40.6 , d ) 40.8 , e ) none of the above | b | divide(subtract(add(multiply(40.2, 10), add(13, 16)), 31), 10) | add(n2,n3)|multiply(n0,n1)|add(#0,#1)|subtract(#2,n4)|divide(#3,n0)| | general | B |
the compound interest on $ 27,000 at 11 % p . a is $ 11,000 . find out the period ( in years ) ? | "amount = 27,000 + 11,000 = 38,000 . . . . . . . . . now by using compound interest formula we get , 27,000 * ( 111 / 100 ) ^ n = 38000 = = = = > n = = 3 1 / 4 years ans - b" | a ) 2.5 , b ) 3 1 / 4 , c ) 3 , d ) 2 3 / 4 , e ) 4 | b | divide(27,000, 11,000) | divide(n0,n2)| | gain | B |
a cargo ships engines failed 100 miles away from the port . due to the changing wind direction , it is moving 12 miles towards the port and 6 miles away from the port . if the wind pattern remains same , how many miles it will travel before reaching the port ? | ships overall distance covered per cycle is + 6 miles take 100 - 6 take 94 / 6 to lowest divisible number - 90 / 6 this means that it will take 15 ` ` overall ' ' actions to reach the 90 th mile . - 15 * 18 miles added later set cycle start at 90 . travel 10 miles and reach the anchor point - add 10 to total 15 * 18 + 10 = 280 miles in total answer : c | a ) a - 179 , b ) b - 240 , c ) c - 280 , d ) d - 100 , e ) e - 155 | c | subtract(add(100, 12), 6) | add(n0,n1)|subtract(#0,n2) | physics | C |
the price of a book is increased from $ 300 to $ 360 . what is the % of increase in its price ? | "explanation : change in the price = rs 360 – rs 300 = rs 60 percentage of increase = change in the price initial price * 100 . percentage increase in price = ( 60 300 ) * 100 = 20 % b" | a ) 10 % , b ) 20 % , c ) 35 % , d ) 40 % , e ) 50 % | b | multiply(divide(subtract(360, 300), 300), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain | B |
if a tire rotates at 400 revolutions per minute when the car is traveling 120 km / h , what is the circumference of the tire ? | 400 rev / minute = 400 * 60 rev / 60 minutes = 24,000 rev / hour 24,000 * c = 12,0000 m : c is the circumference c = 5 meters correct answer d | ['a ) 7 meters', 'b ) 9 meters', 'c ) 8 meters', 'd ) 5 meters', 'e ) 3 meters'] | d | multiply(divide(120, multiply(multiply(multiply(const_2, const_3), const_10), 400)), const_1000) | multiply(const_2,const_3)|multiply(#0,const_10)|multiply(n0,#1)|divide(n1,#2)|multiply(#3,const_1000) | physics | D |
3 friends a , b , c went for week end party to mcdonald ’ s restaurant and there they measure there weights in some order in 7 rounds . a , b , c , ab , bc , ac , abc . final round measure is 165 kg then find the average weight of all the 7 rounds ? | "average weight = [ ( a + b + c + ( a + b ) + ( b + c ) + ( c + a ) + ( a + b + c ) ] / 7 = 4 ( a + b + c ) / 7 = 4 x 165 / 7 = 94.2 kgs answer : a" | a ) 94.2 kgs , b ) 88.5 kgs , c ) 86.5 kgs , d ) 67.5 kgs , e ) 88.2 kgs | a | divide(multiply(add(const_1, 3), 165), 7) | add(const_1,n0)|multiply(n2,#0)|divide(#1,n1)| | general | A |
if d = 1 / ( 2 ^ 3 * 5 ^ 11 ) is expressed as a terminating decimal , how many nonzero digits will d have ? | "another way to do it is : we know x ^ a * y ^ a = ( x * y ) ^ a given = 1 / ( 2 ^ 3 * 5 ^ 11 ) = multiply and divide by 2 ^ 8 = 2 ^ 8 / ( 2 ^ 3 * 2 ^ 8 * 5 ^ 11 ) = 2 ^ 8 / 10 ^ 11 = > non zero digits are 256 = > ans c" | a ) one , b ) two , c ) three , d ) seven , e ) ten | c | add(1, 2) | add(n0,n1)| | general | C |
there were totally 100 men . 82 are married . 75 have t . v , 85 have radio , 70 have a . c . how many men have t . v , radio , a . c and also married ? | "100 - ( 100 - 82 ) - ( 100 - 75 ) - ( 100 - 85 ) - ( 100 - 70 ) = 100 - 18 - 25 - 15 - 30 = 100 - 88 = 12 answer : b" | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | b | subtract(100, add(add(add(subtract(100, 82), subtract(100, 75)), subtract(100, 85)), subtract(100, 70))) | subtract(n0,n1)|subtract(n0,n2)|subtract(n0,n3)|subtract(n0,n4)|add(#0,#1)|add(#4,#2)|add(#5,#3)|subtract(n0,#6)| | general | B |
the cost price of an article is 70 % of the marked price . calculate the gain percent after allowing a discount of 10 % . | "sol . let marked price = rs . 100 . then , c . p . = rs . 70 . s . p = rs . 90 . â ˆ ´ gain % = [ 20 / 70 * 100 ] % = 28.6 % . answer b" | a ) 13.6 % , b ) 28.6 % , c ) 35.6 % , d ) 42.3 % , e ) none | b | multiply(subtract(divide(subtract(const_100, 10), 70), const_1), const_100) | subtract(const_100,n1)|divide(#0,n0)|subtract(#1,const_1)|multiply(#2,const_100)| | gain | B |
along a yard 441 metres long , 22 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between two consecutive trees | "explanation : 22 trees have 21 gaps between them , required distance ( 441 / 21 ) = 21 option d" | a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 12 | d | divide(441, subtract(22, const_1)) | subtract(n1,const_1)|divide(n0,#0)| | physics | D |
let the number which when multiplied by 12 is increased by 198 . | "solution let the number be x . then , 12 x - x = 198 ‹ = › 11 x = 198 x ‹ = › 18 . answer a" | a ) 18 , b ) 20 , c ) 26 , d ) 28 , e ) 30 | a | divide(198, subtract(12, const_1)) | subtract(n0,const_1)|divide(n1,#0)| | general | A |
a , b and c have rs . 300 between them , a and c together have rs . 200 and b and c rs . 350 . how much does c have ? | "a + b + c = 300 a + c = 200 b + c = 350 - - - - - - - - - - - - - - a + b + 2 c = 550 a + b + c = 300 - - - - - - - - - - - - - - - - c = 250 answer : a" | a ) a ) 250 , b ) b ) 249 , c ) c ) 248 , d ) d ) 247 , e ) e ) 246 | a | subtract(add(200, 350), 300) | add(n1,n2)|subtract(#0,n0)| | general | A |
if w / x = 1 / 6 and w / y = 1 / 5 , then ( x + y ) / y = | "w / x = 1 / 6 = > x = 6 w and w / y = 1 / 5 = > y = 5 w ( x + y ) / y = ( 6 w + 5 w ) / 5 w = 11 w / 5 w = 11 / 5 correct option : d" | a ) 4 / 5 , b ) 6 / 5 , c ) 7 / 5 , d ) 11 / 5 , e ) 9 / 5 | d | add(divide(divide(1, 1), divide(5, 6)), const_1) | divide(n2,n0)|divide(n3,n1)|divide(#0,#1)|add(#2,const_1)| | general | D |
if 4 b ^ 4 − 41 b ^ 2 + 100 = 0 , then what is the sum of the two greatest possible values of b ? | just forget about the value of b ^ 4 for the moment and let b ^ 2 = x . ( its better to work with power of 2 than 4 ) now we have equation as 4 x ^ 2 - 41 x + 100 = 0 , factoring them gives us 4 x ^ 2 - 16 x - 25 x + 100 = 0 . which can be solved as ( 4 x - 25 ) ( x - 4 ) , so x = 4 and 25 / 4 . so b ^ 2 = 4 and 25 / 4 so b = + 2 and minus 2 and + 5 / 2 and minus 5 / 2 the two greatest values are + 2 and + 5 / 2 . so their sum 2 + 5 / 2 = 9 / 2 = b | a ) 4 , b ) 9 / 2 , c ) 7 , d ) 41 / 4 , e ) 25 | b | divide(add(add(4, 4), const_1), const_2) | add(n0,n0)|add(#0,const_1)|divide(#1,const_2) | general | B |
speed of a boat in standing water is 9 kmph and the speed of the stream is 1.5 kmph . a man rows to a place at a distance of 105 km and comes back to the starting point . the total time taken by him is : | "speed of upstream = 9 - 1.5 = 7.5 kmph speed of down stream = 9 + 1.5 = 10.5 kmph total time = ( 105 / 7.5 ) + ( 105 / 10.5 ) = 24 hrs answer : d" | a ) 16 hours , b ) 18 hours , c ) 20 hours , d ) 24 hours , e ) 26 hours | d | add(divide(105, add(9, 1.5)), divide(105, subtract(9, 1.5))) | add(n0,n1)|subtract(n0,n1)|divide(n2,#0)|divide(n2,#1)|add(#2,#3)| | physics | D |
a train running at the speed of 60 km / hr crosses a pole in 36 seconds . what is the length of the train ? | "speed = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec length of the train = ( speed x time ) = ( 50 / 3 * 36 ) m = 600 m . answer : d" | a ) 186 m , b ) 176 m , c ) 100 m , d ) 600 m , e ) 765 m | d | multiply(divide(multiply(60, const_1000), const_3600), 36) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics | D |
find the missing figures : 0.25 % of ? = 0.04 | "let 0.25 % of x = 0.04 . then , 0.25 * x / 100 = 0.04 x = [ ( 0.04 * 100 ) / 0.25 ] = 16 . answer is a ." | a ) 16 , b ) 20 , c ) 17 , d ) 12 , e ) 18 | a | divide(0.04, divide(0.25, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain | A |
if 5 machines can produce 20 units in 10 hours , how long would it take 20 machines to produce 200 units ? | here , we ' re told that 5 machines can produce 20 units in 10 hours . . . . that means that each machine works for 10 hours apiece . since there are 5 machines ( and we ' re meant to assume that each machine does the same amount of work ) , then the 5 machines equally created the 20 units . 20 units / 5 machines = 4 units are made by each machine every 10 hours now that we know how long it takes each machine to make 4 units , we can break this down further if we choose to . . . 10 hours / 4 units = 2.5 hours per unit when 1 machine is working . the prompt asks us how long would it take 20 machines to produce 200 units . if 20 machines each work for 2.5 hours , then we ' ll have 20 units . since 200 units is ' 10 times ' 20 , we need ' 10 times ' more time . ( 2.5 hours ) ( 10 times ) = 25 hours final answer : [ reveal ] spoiler : c | a ) 50 hours , b ) 40 hours , c ) 25 hours , d ) 12 hours , e ) 8 hours | c | divide(200, multiply(divide(divide(20, 10), 5), 20)) | divide(n1,n2)|divide(#0,n0)|multiply(n1,#1)|divide(n4,#2) | physics | C |
a work can be finished in 16 days by 30 women . the same work can be finished in 10 days by sixteen men . the ratio between the capacity of a man and a woman is | "work done by 30 women in 1 day = 1 / 16 work done by 1 woman in 1 day = 1 / ( 16 × 30 ) work done by 16 men in 1 day = 1 / 10 work done by 1 man in 1 day = 1 / ( 10 × 16 ) ratio of the capacity of a man and woman = 1 / ( 10 × 16 ) : 1 / ( 16 × 30 ) = 1 / 10 : 1 / 30 = 1 / 1 : 1 / 3 = 3 : 1 option a" | a ) 3 : 1 , b ) 4 : 3 , c ) 2 : 3 , d ) 2 : 1 , e ) 2 : 4 | a | 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 |
a polling company surveyed a certain country , and it found that 35 % of that country ’ s registered voters had an unfavorable impression of both of that state ’ s major political parties and that 20 % had a favorable impression only of party e . if one registered voter has a favorable impression of both parties for every two registered voters who have a favorable impression only of party b , then what percentage of the country ’ s registered voters have a favorable impression of both parties ( assuming that respondents to the poll were given a choice between favorable and unfavorable impressions only ) ? | "s = 100 not ( e and b ) = 35 only e = 20 ( e and b ) / b = 1 / 2 let ( e and b ) = x only b = 2 x so now , 20 + 35 + x + 2 x = 100 x = 15 a ans" | a ) 15 , b ) 20 , c ) 30 , d ) 35 , e ) 45 | a | divide(subtract(const_100, add(35, 20)), const_3) | add(n0,n1)|subtract(const_100,#0)|divide(#1,const_3)| | gain | A |
a reduction of 50 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 800 , what is the reduced price for kg ? | "800 * ( 50 / 100 ) = 400 - - - - 5 ? - - - - 1 = > rs . 80 answer : d" | a ) s . 42 , b ) s . 46 , c ) s . 40 , d ) s . 80 , e ) s . 34 | d | divide(divide(multiply(800, 50), const_100), 5) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)| | gain | D |
in 1982 and 1983 , company b ’ s operating expenses were $ 13.0 million and $ 15.0 million , respectively , and its revenues were $ 15.6 million and $ 18.8 million , respectively . what was the percent increase in company b ’ s profit ( revenues minus operating expenses ) from 1982 to 1983 ? | "profit in 1982 = 15.6 - 13 = 2.6 million $ profit in 1983 = 18.8 - 15 = 3.8 million $ percentage increase in profit = ( 3.8 - 2.6 ) / 2.6 * 100 % = 46 1 / 6 % answer b" | a ) 3 % , b ) 46 1 / 6 % , c ) 25 % , d ) 33 1 / 3 % , e ) 60 % | b | subtract(const_100, divide(multiply(subtract(15.0, 13.0), const_100), subtract(18.8, 15.6))) | subtract(n3,n2)|subtract(n5,n4)|multiply(#0,const_100)|divide(#2,#1)|subtract(const_100,#3)| | general | B |
a hiker walked for 3 days . she walked 18 miles on the first day , walking 3 miles per hour . on the second day she walked for one less hour but she walked one mile per hour , faster than on the first day . on the third day she walked the same number of hours as on the first day , but at the same speed as on the second day . how many miles in total did she walk ? | first day - 18 miles with 3 miles per hours then total - 6 hours for that day second day - 4 miles per hour and 5 hours - 20 miles third day - 4 miles per hour and 6 hours - 24 miles total 18 + 20 + 24 - 62 answer : option e . | a ) 24 , b ) 44 , c ) 58 , d ) 60 , e ) 62 | e | add(add(18, multiply(add(3, const_1), subtract(divide(18, 3), const_1))), multiply(divide(18, 3), add(3, const_1))) | add(n0,const_1)|divide(n1,n0)|multiply(#0,#1)|subtract(#1,const_1)|multiply(#0,#3)|add(n1,#4)|add(#5,#2) | physics | E |
two trains of equal length , running with the speeds of 60 and 40 kmph , take 36 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ? | rs = 60 - 40 = 20 * 5 / 18 = 100 / 18 t = 36 d = 36 * 100 / 18 = 200 rs = 60 + 40 = 100 * 5 / 18 t = 200 * 18 / 500 = 7.2 sec answer : d | a ) 10 sec , b ) 16 sec , c ) 19 sec , d ) 7.2 sec , e ) 13 sec | d | multiply(multiply(multiply(const_0_2778, subtract(60, 40)), 36), inverse(multiply(const_0_2778, add(60, 40)))) | add(n0,n1)|subtract(n0,n1)|multiply(#0,const_0_2778)|multiply(#1,const_0_2778)|inverse(#2)|multiply(n2,#3)|multiply(#4,#5) | physics | D |
if 1 / 2 of the air in a tank is removed with each stroke of a vacuum pump , what fraction of the original amount of air has been removed after 2 strokes ? | "left after 1 st stroke = 1 / 2 left after 2 nd stroke = 1 / 2 * 1 / 2 = 1 / 4 so removed = 1 - 1 / 4 = 3 / 4" | a ) 3 / 4 , b ) 7 / 8 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 16 | a | add(add(add(add(divide(1, 2), divide(divide(1, 2), 2)), divide(divide(divide(1, 2), 2), 2)), divide(divide(divide(divide(1, 2), 2), 2), 2)), divide(divide(divide(divide(divide(1, 2), 2), 2), 2), 2)) | divide(n0,n1)|divide(#0,n1)|add(#0,#1)|divide(#1,n1)|add(#2,#3)|divide(#3,n1)|add(#4,#5)|divide(#5,n1)|add(#6,#7)| | physics | A |
how many positive integers less than 100 are neither multiples of 4 or 5 . | "to answer this q we require to know 1 ) multiples of 4 till 100 = 100 / 4 = 25 2 ) multiples of 5 till 100 = 100 / 5 = 20 add the two 25 + 20 = 45 ; subtract common terms that are multiple of both 4 and 5 . . lcm of 4 and 5 = 20 multiples of 20 till 100 = 100 / 20 = 5 = so total multiples of 4 and 5 = 45 - 5 = 40 ans = 100 - 40 = 60 b" | a ) a ) 30 , b ) b ) 60 , c ) c ) 32 , d ) d ) 33 , e ) e ) 34 | b | divide(factorial(subtract(add(const_4, 4), const_1)), multiply(factorial(4), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general | B |
the average weight of 12 person ' s increases by 4 kg when a new person comes in place of one of them weighing 58 kg . what is the weight of the new person ? | "total increase in weight = 12 × 4 = 48 if x is the weight of the new person , total increase in weight = x − 58 = > 48 = x - 58 = > x = 48 + 58 = 106 answer : e" | a ) 85 , b ) 80 , c ) 95 , d ) 90 , e ) 106 | e | add(multiply(12, 4), 58) | multiply(n0,n1)|add(n2,#0)| | general | E |
the h . c . f . of two numbers is 23 and the other two factors of their l . c . m . are 13 and 17 . the larger of the two numbers is : | "clearly , the numbers are ( 23 x 13 ) and ( 23 x 17 ) . larger number = ( 23 x 17 ) = 391 . answer : option c" | a ) 276 , b ) 299 , c ) 391 , d ) 345 , e ) 355 | c | multiply(23, 17) | multiply(n0,n2)| | other | C |
a train is travelling at 48 kmph . it crosses another train having half of its length , travelling in opposite direction at 42 kmph , in 12 seconds . it also passes a railway platform in 45 seconds . what is the length of the platform ? | speed of i train = 48 kmph length of i train = 2 x m speed of sec train = 42 kmph length of seconf train = x m dist = ( 2 x + x ) = 3 x m time = 12 sec = 90 * 5 / 18 = 25 m / s x = 100 length of first train = 2 x = 200 m time taken = 45 sec speed of first train = 48 kmph = 48 * 5 / 18 = 40 / 3 m / s length of the platform = y metre 200 + y = 45 * 45 / 3 = 600 y = 400 answer a | a ) 400 m , b ) 500 m , c ) 600 m , d ) 550 m , e ) 700 m | a | subtract(multiply(multiply(45, 48), const_0_2778), divide(multiply(multiply(add(48, 42), const_0_2778), 12), divide(const_3, const_2))) | add(n0,n1)|divide(const_3,const_2)|multiply(n0,n3)|multiply(#2,const_0_2778)|multiply(#0,const_0_2778)|multiply(n2,#4)|divide(#5,#1)|subtract(#3,#6) | physics | A |
for each of her sales , a saleswoman receives a commission equal to 20 percent of the first $ 500 of the total amount of the sale , plus 40 percent of the total amount in excess of $ 500 . if the total amount of one of her sales was $ 800 , the saleswoman ’ s commission was approximately what percent of the total amount of the sale ? | "total sales = 800 comission = ( 20 / 100 ) * 500 + ( 40 / 100 ) * 300 = 100 + 120 = 220 % comission = ( 220 / 800 ) * 100 = 27.5 ~ 28 % answer is e" | a ) 22 % , b ) 24 % , c ) 25 % , d ) 27 % , e ) 28 % | e | multiply(divide(add(multiply(divide(20, const_100), 500), multiply(divide(40, const_100), subtract(800, 500))), 800), const_100) | divide(n0,const_100)|divide(n2,const_100)|subtract(n4,n1)|multiply(n1,#0)|multiply(#1,#2)|add(#3,#4)|divide(#5,n4)|multiply(#6,const_100)| | general | E |
the sum of the present age of henry and jill is 48 . what is their present ages if 9 years ago henry was twice the age of jill ? | "let the age of jill 9 years ago be x , age of henry be 2 x x + 9 + 2 x + 9 = 48 x = 10 present ages will be 19 and 29 answer : d" | a ) and 27 , b ) and 24 , c ) and 22 , d ) and 29 , e ) of these | d | subtract(48, divide(add(48, 9), const_3)) | add(n0,n1)|divide(#0,const_3)|subtract(n0,#1)| | general | D |
the volume of the largest right circular cone that can be cut out of a cube of edge 7 cm is : | sol . volume of the block = ( 10 * 5 * 2 ) cm ³ . volume of the cone carved out = [ 1 / 3 * 22 / 7 * 3 * 3 * 7 ] cm ³ = 66 cm ³ ∴ wood wasted = ( 100 - 66 ) % = 34 % . answer a | ['a ) 89.8 cm ³', 'b ) 92.5 cm ³', 'c ) 132.8 cm ³', 'd ) 144.5 cm ³', 'e ) none'] | a | divide(multiply(multiply(multiply(multiply(const_1, add(multiply(7, const_3), const_1)), divide(7, const_2)), divide(7, const_2)), 7), multiply(const_3, 7)) | divide(n0,const_2)|multiply(n0,const_3)|add(#1,const_1)|multiply(#2,const_1)|multiply(#0,#3)|multiply(#0,#4)|multiply(n0,#5)|divide(#6,#1) | geometry | A |
if 20 % of a is the same as 25 % of b , then a : b is : | "expl : 20 % of a i = 25 % of b = 20 a / 100 = 25 b / 100 = 5 / 4 = 5 : 4 answer : c" | a ) 3 : 4 , b ) 4 : 3 , c ) 5 : 4 , d ) 6 : 7 , e ) 5 : 7 | c | divide(divide(25, const_100), divide(20, const_100)) | divide(n1,const_100)|divide(n0,const_100)|divide(#0,#1)| | gain | C |
the difference between the place values of 8 and 3 in the number 587435 is | sol . = ( place value of 8 ) – ( place value of 3 ) = ( 80000 - 30 ) = 79970 answer b | a ) 75200 , b ) 79970 , c ) 75460 , d ) 24580 , e ) none | b | subtract(multiply(const_10, 8), 8) | multiply(n0,const_10)|subtract(#0,n0)| | general | B |
the average height of 30 students in a class was calculated as 175 cm . it has later found that the height of one of the students in the class was incorrectly written as 151 cm whereas the actual height was 136 cm . what was the actual average height of the students in the class ? | "the total height was 15 cm too much . the average height should be reduced by 15 cm / 30 = 0.5 cm the answer is a ." | a ) 174.5 , b ) 173.5 , c ) 172.5 , d ) 171.5 , e ) 170.5 | a | divide(subtract(multiply(30, 175), subtract(151, 136)), 30) | multiply(n0,n1)|subtract(n2,n3)|subtract(#0,#1)|divide(#2,n0)| | general | A |
at a special sale , 8 tickets can be purchased for the price of 6 tickets . if 8 tickets are purchased at the sale , the amount saved will be what percent of the original price of the 8 tickets ? | "let the price of a ticket be rs . 100 , so 6 tickets cost 600 & 8 tickets cost 800 8 tickets purchased at price of 6 tickets ie . , for 600 , so amount saved s rs . 200 , % of 6 tickets = ( 200 / 800 ) * 100 = 25 % answer : a" | a ) 25 % , b ) 33.3 % , c ) 40 % , d ) 60 % , e ) 66.6 % | a | divide(multiply(subtract(multiply(8, 8), multiply(6, 8)), const_100), multiply(8, 8)) | multiply(n0,n0)|multiply(n0,n1)|subtract(#0,#1)|multiply(#2,const_100)|divide(#3,#0)| | gain | A |
in what time will a railway train 70 m long moving at the rate of 36 kmph pass a telegraph post on its way ? | "t = 70 / 36 * 18 / 5 = 7 sec answer : b" | a ) 6 sec , b ) 7 sec , c ) 8 sec , d ) 9 sec , e ) 2 sec | b | divide(70, multiply(36, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics | B |
if x + y = 9 and x – y = 3 , then x ^ 2 - y ^ 2 = | the fastest approach has already been shown . here ' s one more option . given : x + y = 9 x – y = 3 add the two equations to get : 2 x = 12 , which means x = 6 if x = 6 , we can plug that value into either equation , to conclude that y = 3 if x = 6 and y = 3 , then x ² - y ² = 6 ² - 3 ² = 27 answer : d | a ) - 4 , b ) 4 , c ) 10 , d ) 27 , e ) 40 | d | subtract(power(subtract(9, 3), 2), power(const_3, const_2)) | power(const_3,const_2)|subtract(n0,n1)|power(#1,n2)|subtract(#2,#0) | general | D |
a merchant marks goods up by 75 % and then offers a discount on the marked price . the profit that the merchant makes after offering the discount is 5 % . what % discount did the merchant offer ? | "let p be the original price of the goods and let x be the rate after the markup . ( 1.75 p ) * x = 1.05 p x = 1.05 / 1.75 = 0.6 which is a discount of 40 % . the answer is d ." | a ) 20 % , b ) 25 % , c ) 35 % , d ) 40 % , e ) 45 % | d | multiply(divide(subtract(subtract(add(const_1, divide(75, const_100)), const_1), divide(5, const_100)), add(const_1, divide(75, const_100))), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(#2,const_1)|subtract(#3,#1)|divide(#4,#2)|multiply(#5,const_100)| | gain | D |
alok ordered 16 chapatis , 5 plates of rice , 7 plates of mixed vegetable and 6 ice - cream cups . the cost of each chapati is rs . 6 , that of each plate of rice is rs . 45 and that of mixed vegetable is rs . 70 . the amount that alok paid the cashier was rs . 1015 . find the cost of each ice - cream cup ? | "explanation : let the cost of each ice - cream cup be rs . x 16 ( 6 ) + 5 ( 45 ) + 7 ( 70 ) + 6 ( x ) = 1015 96 + 225 + 490 + 6 x = 1015 6 x = 204 = > x = 34 . answer : a" | a ) 34 , b ) 76 , c ) 29 , d ) 12 , e ) 20 | a | divide(subtract(subtract(subtract(1015, multiply(16, 6)), multiply(5, 45)), multiply(7, 70)), 6) | multiply(n0,n3)|multiply(n1,n5)|multiply(n2,n6)|subtract(n7,#0)|subtract(#3,#1)|subtract(#4,#2)|divide(#5,n3)| | general | A |
during the first two weeks of january , the total rainfall in springdale was 30 inches . if the rainfall during the second week was 1.5 times the rainfall during the first week , what was the rainfall during the second week of january ? | "total rainfall in 2 weeks = 18 inches . assume the rainfall in second week = 1 . x rainfall in first week = x total rainfall = 2.5 x = 18 inches x = 12 and 1.5 x = 18 rainfall during second week = 18 inches option e" | a ) 5 inches , b ) 6 inches , c ) 9 inches , d ) 10 inches , e ) 18 inches | e | multiply(30, divide(1.5, add(const_1, 1.5))) | add(n1,const_1)|divide(n1,#0)|multiply(n0,#1)| | general | E |
log 3 n + log 5 n what is 3 digit number n that will be whole number | "no of values n can take is 1 5 ^ 3 = 125 answer : d" | a ) 629 , b ) 729 , c ) 829 , d ) 125 , e ) 727 | d | power(5, 3) | power(n1,n0)| | other | D |
there are 300 seniors at morse high school , and 40 % of them have cars . of the remaining grades ( freshmen , sophomores , and juniors ) , only 10 % of them have cars . if 15 % of all the students at morse have cars , how many students are in those other 3 lower grades ? | let the total no of students be n . the no of seniors having cars is 40 % of 300 i . e . 120 . the rest of the students with other three grades who have cars is 10 % of ( n - 300 ) . the total no of students with cars is 15 % of n . thus , 0.15 n = 120 + 0.1 ( n - 300 ) on solving this , we get n = 1800 . hence , the no of students with other three grades is 1800 - 300 , i . e . 1500 ( e ) . | a ) 600 , b ) 900 , c ) 1200 , d ) 1350 , e ) 1500 | e | multiply(15, const_100) | multiply(n3,const_100) | gain | E |
when you draw 2 dices together , find the probability of getting a total of 6 ? | s = { ( 1,1 ) , ( 1,2 ) , ( 1,3 ) , ( 1,4 ) , ( 1,5 ) , ( 1,6 ) , ( 4,1 ) , ( 4,2 ) , ( 4,3 ) , ( 4,4 ) , ( 4,5 ) , ( 4,6 ) , ( 2,1 ) , ( 2,2 ) , ( 2,3 ) , ( 2,4 ) , ( 2,5 ) , ( 2,6 ) , ( 5,1 ) , ( 5,2 ) , ( 5,3 ) , ( 5,4 ) , ( 5,5 ) , ( 5,6 ) , ( 3,1 ) , ( 3,2 ) , ( 3,3 ) , ( 3,4 ) , ( 3,5 ) , ( 3,6 ) , ( 6,1 ) , ( 6,2 ) , ( 6,3 ) , ( 6,4 ) , ( 6,5 ) , ( 6,6 ) } n ( s ) = 36 set of total of 6 , a = { ( 1,5 ) , ( 5,1 ) , ( 2,4 ) , ( 4,2 ) , ( 3,3 ) } ; n ( a ) = 5 p ( total of 6 ) = n ( a ) n ( s ) n ( a ) n ( s ) = 5 / 36 c | a ) 3 / 13 , b ) 2 / 31 , c ) 5 / 36 , d ) 7 / 41 , e ) 7 / 43 | c | add(divide(2, power(6, 2)), divide(const_3, power(6, 2))) | power(n1,n0)|divide(n0,#0)|divide(const_3,#0)|add(#1,#2) | probability | C |
the area of a triangle is with base 3 m and height 6 m ? | "1 / 2 * 3 * 6 = 9 m 2 answer : b" | a ) 88 m 2 , b ) 9 m 2 , c ) 66 m 2 , d ) 77 m 2 , e ) 31 m 2 | b | triangle_area(3, 6) | triangle_area(n0,n1)| | geometry | B |
a survey of n people in the town of eros found that 50 % of them preferred brand a . another survey of 130 people in the town of angie found that 60 % preferred brand a . in total , 55 % of all the people surveyed together preferred brand a . what is the total number of people surveyed ? | "it is simply a weighted average question . since the given average of 50 % and 60 % is 55 % ( right in the middle ) , it means the number of people surveyed in eros ( n ) is same as the number of people surveyed in angie . so n = 130 total = 130 + 130 = 260 answer ( d )" | a ) 50 , b ) 100 , c ) 150 , d ) 260 , e ) 250 | d | divide(subtract(multiply(130, divide(60, const_100)), multiply(130, divide(55, const_100))), subtract(divide(55, const_100), divide(50, const_100))) | divide(n2,const_100)|divide(n3,const_100)|divide(n0,const_100)|multiply(n1,#0)|multiply(n1,#1)|subtract(#1,#2)|subtract(#3,#4)|divide(#6,#5)| | general | D |
if the ratio of two number is 5 : 7 and lcm of the number is 320 then what is the number . | "product of two no = lcm * hcf 5 x * 7 x = 320 * x x = 9 answer : b" | a ) 15 , b ) 9 , c ) 10 , d ) 11 , e ) 35 | b | divide(320, multiply(5, 7)) | multiply(n0,n1)|divide(n2,#0)| | other | B |
a flagstaff 17.5 m high casts a shadow of length 40.25 m . the height of the building , which casts a shadow of length 28.75 m under similar conditions will be | "let the height of the building x metres . less lengthy shadow , less in the height ( direct proportion ) 40.25 : 28.75 : : 17.5 : x 40.25 * x = 28.75 * 17.5 x = ( 28.75 x 17.5 ) / 40.25 x = 12.5 answer is a ." | a ) 12.5 m , b ) 10 m , c ) 17.5 m , d ) 21.25 m , e ) 9.8 m | a | multiply(28.75, divide(17.5, 40.25)) | divide(n0,n1)|multiply(n2,#0)| | physics | A |
according to the direction on a can of frozen orange juice concentrate is to be mixed with 3 cans of water to make orange juice . how many 15 - ounce cans of the concentrate are required to prepare 200 6 - ounce servings of orange juice ? | "orange juice concentrate : water : : 1 : 3 total quantity of orange juice = 200 * 6 = 1200 oz so orange juice concentrate : water : : 300 oz : 900 oz no . of 15 oz can = 300 oz / 15 oz = 20 answer d , 20 cans" | a ) 25 , b ) 34 , c ) 50 , d ) 20 , e ) 100 | d | divide(multiply(6, 200), divide(200, const_10)) | divide(n2,const_10)|multiply(n2,n3)|divide(#1,#0)| | general | D |
( 4300731 ) - ? = 2535618 | "let 4300731 - x = 2535618 then x = 4300731 - 2535618 = 1765113 answer is c" | a ) 1865113 , b ) 1775123 , c ) 1765113 , d ) 1675123 , e ) none of them | c | multiply(4300731, power(add(const_4, const_1), const_4)) | add(const_1,const_4)|power(#0,const_4)|multiply(n0,#1)| | general | C |
an amount of money is to be distributed among faruk , vasim and ranjith in the ratio 3 : 5 : 9 . if vasims share is rs . 1500 , what is the difference between faruk ' s and ranjith ' s shares ? | "explanation : let p = faruk , q = vasim , r = ranjith let p = 3 x , q = 5 x and r = 9 x . then , 5 x = 1500 ? x = 300 . p = 900 , q = 1500 and r = 2700 . hence , ( r - p ) = ( 2700 - 900 ) = 1800 answer : a" | a ) s 1800 , b ) s 1500 , c ) s 1600 , d ) s 1900 , e ) s 1700 | a | multiply(divide(1500, 5), subtract(9, 3)) | divide(n3,n1)|subtract(n2,n0)|multiply(#0,#1)| | other | A |
if you wish to live to be 100 years old ( hypothetically ) , you must consume 500 calories less than your average daily allowance for your age . if you are in your 60 ' s , and your average daily allowance is 2000 calories per day , how many calories are you allowed in a week ? | to determine the amount of calories allowed per day , you need to subtract 500 from 2000 , for a total of 1500 to determine the amount of calories allowed per week , you need to multiply the daily amount allowed ( 1500 ) by the number of days in a week ( 7 ) for a total of 10500 the correct answer is b | a ) 7000 , b ) 10500 , c ) 14000 , d ) 12000 , e ) 3500 | b | multiply(subtract(2000, 500), add(const_4, const_3)) | add(const_3,const_4)|subtract(n3,n1)|multiply(#0,#1) | general | B |
a man engaged a servant on the condition that he would pay him rs . 800 and a uniform after one year service . he served only for 9 months and received uniform and rs . 400 , find the price of the uniform ? | "9 / 12 = 3 / 4 * 800 = 600 400 - - - - - - - - - - - - - 200 1 / 4 - - - - - - - - 200 1 - - - - - - - - - ? = > rs . 800 answer : a" | a ) s . 800 , b ) s . 85 , c ) s . 90 , d ) s . 100 , e ) s . 120 | a | multiply(divide(subtract(multiply(9, 800), multiply(multiply(const_3, const_4), 400)), 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 |
what is the total number of prime numbers less than 70 ? | solution prime numbers less than 70 are = 12 . 2,3 , 5,7 , 11,13 , 17,19 , 23,29 , 31,37 , 41,43 , 47,53 , 59,61 , and 67 their number is 19 . answer c | a ) 17 , b ) 18 , c ) 19 , d ) 20 , e ) 21 | c | max(add(add(const_10, const_3), const_4), add(add(add(const_10, const_3), const_4), const_2)) | add(const_10,const_3)|add(#0,const_4)|add(#1,const_2)|max(#1,#2) | general | C |
a man spend 1 / 5 of his salary on food , 1 / 10 of his salary on house rent and 3 / 5 salary on clothes . he still has $ 17000 left with him . find salary . . | "[ 1 / ( x 1 / y 1 + x 2 / y 2 + x 3 / y 3 ) ] * total amount = balance amount [ 1 - ( 1 / 5 + 1 / 10 + 3 / 5 ) } * total salary = $ 17000 , = [ 1 - 9 / 10 ] * total salary = $ 17000 , total salary = $ 17000 * 10 = $ 180000 , correct answer ( e )" | a ) $ 10700 , b ) $ 17000 , c ) $ 180000 , d ) $ 1700 , e ) $ 170000 | e | divide(17000, subtract(1, add(add(divide(1, 5), divide(1, 10)), divide(3, 5)))) | divide(n0,n1)|divide(n0,n3)|divide(n4,n1)|add(#0,#1)|add(#3,#2)|subtract(n0,#4)|divide(n6,#5)| | general | E |
a number when divided by 342 gives a remainder 47 . when the same number if it divided by 19 , what would be the remainder ? | "on dividing the given number by 342 , let k be the quotient and 47 as remainder . then , number – 342 k + 47 = ( 19 x 18 k + 19 x 2 + 9 ) = 19 ( 18 k + 2 ) + 9 . the given number when divided by 19 , gives ( 18 k + 2 ) as quotient and 9 as remainder . answer d" | a ) 1 , b ) 2 , c ) 5 , d ) 9 , e ) 11 | d | subtract(47, multiply(19, const_2)) | multiply(n2,const_2)|subtract(n1,#0)| | general | D |
a , b and c rent a pasture . if a puts 10 oxen for 7 months , b puts 12 oxen for 5 months and c puts 15 oxen for 3 months for grazing and the rent of the pasture is rs . 175 , then how much amount should c pay as his share of rent ? | "a : b : c = 10 × 7 : 12 × 5 : 15 × 3 = 2 × 7 : 12 × 1 : 3 × 3 = 14 : 12 : 9 amount that c should pay = 175 × ( 9 / 35 ) = 5 × 9 = 45 answer is b ." | a ) 40 , b ) 45 , c ) 49 , d ) 51 , e ) 43 | b | multiply(175, divide(multiply(15, 3), add(add(multiply(10, 7), multiply(12, 5)), multiply(15, 3)))) | multiply(n4,n5)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|add(#3,#0)|divide(#0,#4)|multiply(n6,#5)| | general | B |
in a factory , there are 30 % technicians and 70 % non - technicians . if the 30 % of the technicians and 70 % of non - technicians are permanent employees , then the percentage of workers who are temporary is ? | "total = 100 t = 30 nt = 70 30 * ( 70 / 100 ) = 21 70 * ( 30 / 100 ) = 21 21 + 21 = 42 = > 100 - 42 = 58 % answer : e" | a ) 62 % , b ) 57 % , c ) 52 % , d ) 22 % , e ) 48 % | e | subtract(const_100, add(multiply(30, divide(70, const_100)), multiply(divide(30, const_100), 70))) | divide(n1,const_100)|divide(n0,const_100)|multiply(n0,#0)|multiply(n1,#1)|add(#2,#3)|subtract(const_100,#4)| | gain | E |
if 213 × 16 = 3408 , then 1.6 × 21.3 is equal to : | "solution 1.6 × 21.3 = ( 16 / 10 x 213 / 10 ) = ( 16 x 213 / 100 ) = 3408 / 100 = 34.08 . answer c" | a ) 0.3408 , b ) 3.408 , c ) 34.08 , d ) 340.8 , e ) none of these | c | multiply(1.6, 21.3) | multiply(n4,n3)| | general | C |
if takes 25.5 minutes to download a file using modem a . modem b work at 17 % of the speed of modem a . how long does it take to download a file using a modem b ? | let speed of modem a be k so speed of modem b is 17 / 100 * k time taken by modem a = 25.5 minutes let time taken by modem b = x k * 25.5 = 17 / 100 * k * x 0.17 * x = 25.5 x = 25.5 / 0.17 = 150 minutes answer : c | a ) 130 minutes , b ) 140 minutes , c ) 150 minutes , d ) 160 minutes , e ) 170 minutes | c | divide(multiply(25.5, const_100), 17) | multiply(n0,const_100)|divide(#0,n1) | physics | C |
a shirt goes on sale for 80 % of its original price . one week later , the sale price is marked down 10 % . the final price is what percent of the original price ? | "just assume original price is 100 . sale price = 80 then it is marked down by 10 % = 80 - 8 = 72 . hence it is 72 % of the original price . hence answer is a" | a ) 72 % , b ) 70 % , c ) 52 % , d ) 50 % , e ) 28 % | a | subtract(80, multiply(divide(80, const_100), 10)) | divide(n0,const_100)|multiply(n1,#0)|subtract(n0,#1)| | gain | A |
of 30 applicants for a job , 13 had at least 4 years ' experience , 18 had degrees , and 3 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | "d . 6 30 - 3 = 27 27 - 13 - 18 = - 6 then 6 are in the intersection between 4 years experience and degree . answer d" | a ) 14 , b ) 13 , c ) 9 , d ) 6 , e ) 5 | d | add(subtract(add(13, 18), subtract(30, 3)), subtract(18, 13)) | add(n1,n3)|subtract(n0,n4)|subtract(n3,n1)|subtract(#0,#1)|add(#3,#2)| | general | D |
the unit digit in the product 1457 * 7863 * 4575 * 9719 is ? | "unit digit in the given product = unit digit in 7 * 3 * 5 * 9 = 5 answer is a" | a ) 5 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | subtract(multiply(multiply(multiply(1457, 7863), 4575), 9719), subtract(multiply(multiply(multiply(1457, 7863), 4575), 9719), add(const_4, const_4))) | add(const_4,const_4)|multiply(n0,n1)|multiply(n2,#1)|multiply(n3,#2)|subtract(#3,#0)|subtract(#3,#4)| | general | A |
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 | C |
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