Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
values |
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60 % of a number is added to 150 , the result is the same number . find the number ? | ": ( 60 / 100 ) * x + 150 = x 2 x = 750 x = 375 answer : a" | a ) 375 , b ) 277 , c ) 266 , d ) 99 , e ) 121 | a | divide(150, divide(150, const_100)) | divide(n1,const_100)|divide(n1,#0)| | gain |
x , y , and z are different prime numbers . the product x ^ 3 * y ^ 2 * z ^ 2 is divisible by how many different positive numbers ? | "the exponents of x ^ 3 * y ^ 2 * z ^ 2 are 3 , 2 , and 2 . the number of factors is ( 3 + 1 ) ( 2 + 1 ) ( 2 + 1 ) = 36 the answer is b ." | a ) 24 , b ) 36 , c ) 48 , d ) 60 , e ) 72 | b | subtract(power(3, const_4), const_4) | power(n0,const_4)|subtract(#0,const_4)| | general |
what is the smallest integer that is multiple of 87 and 20 | correct answer : e it is the lcm of 8,7 and 20 which is 280 | a ) 141 , b ) 180 , c ) 130 , d ) 122 , e ) 280 | e | multiply(divide(lcm(87, 20), const_12), const_2) | lcm(n0,n1)|divide(#0,const_12)|multiply(#1,const_2) | general |
two employees x and y are paid a total of rs . 600 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 600 but x = 120 % of y = 120 y / 100 = 12 y / 10 ∴ 12 y / 10 + y = 600 ⇒ y [ 12 / 10 + 1 ] = 600 ⇒ 22 y / 10 = 600 ⇒ 22 y = 6000 ⇒ y = 6000 / 22 = rs . 272.72 c" | a ) s . 200.72 , b ) s . 250.72 , c ) s . 272.72 , d ) s . 300.72 , e ) s . 310.72 | c | divide(multiply(600, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | add(const_1,const_4)|multiply(#0,const_2)|add(#1,const_1)|multiply(n0,#1)|multiply(#2,const_2)|divide(#3,#4)| | general |
a reduction of 20 % 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 * ( 20 / 100 ) = 160 - - - - 5 ? - - - - 1 = > rs . 32 answer : e" | a ) s . 42 , b ) s . 46 , c ) s . 40 , d ) s . 41 , e ) s . 32 | e | divide(divide(multiply(800, 20), const_100), 5) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)| | gain |
mary ' s income is 40 percent more than tim ' s income , and tim ' s income is 40 percent less than juan ' s income . what percent of juan ' s income is mary ' s income ? | "juan ' s income = 100 ( assume ) ; tim ' s income = 60 ( 40 percent less than juan ' s income ) ; mary ' s income = 84 ( 40 percent more than tim ' s income ) . thus , mary ' s income ( 84 ) is 84 % of juan ' s income ( 100 ) . answer : e ." | a ) 124 % , b ) 120 % , c ) 96 % , d ) 80 % , e ) 84 % | e | multiply(multiply(subtract(const_1, divide(40, const_100)), add(const_1, divide(40, const_100))), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)| | general |
kim has 40 percent more money than sal and sal has 20 percent less money than phil . if sal and phil have a combined total of $ 1.80 , how much money does kim have ? | phil = x sal = 0.8 x kim = ( 1.4 ) * 0.8 x = 1.12 x x + 0.8 x = 1.8 - - > x = 1 kim = 1.12 answer : b | a ) $ 1.00 , b ) $ 1.12 , c ) $ 1.20 , d ) $ 1.32 , e ) $ 1.40 | b | multiply(divide(1.8, add(const_100, subtract(const_100, 20))), divide(multiply(subtract(const_100, 20), add(const_100, 40)), const_100)) | add(n0,const_100)|subtract(const_100,n1)|add(#1,const_100)|multiply(#0,#1)|divide(n2,#2)|divide(#3,const_100)|multiply(#4,#5) | general |
the sides of a rectangle are in the ratio 4 : 3 and its area is 972 sq . m find the perimeter of rectangle | explanation : let 4 x and 3 x be sides of the rectangle we know that area of rectangle = l × b 4 x × 3 x = 972 12 x 2 = 972 x 2 = 81 x = 9 therefore length = 4 x = 4 × 9 = 36 m breadth = 3 x = 3 × 9 = 27 m therefore perimeter = 2 ( l + b ) = 2 ( 36 + 27 ) = 126 m answer : option d | ['a ) 120 m', 'b ) 122 m', 'c ) 124 m', 'd ) 126 m', 'e ) 128 m'] | d | rectangle_perimeter(multiply(4, sqrt(divide(972, multiply(4, 3)))), multiply(3, sqrt(divide(972, multiply(4, 3))))) | multiply(n0,n1)|divide(n2,#0)|sqrt(#1)|multiply(n0,#2)|multiply(n1,#2)|rectangle_perimeter(#3,#4) | geometry |
a gambler bought $ 3,000 worth of chips at a casino in denominations of $ 20 and $ 100 . that evening , the gambler lost 14 chips , and then cashed in the remainder . if the number of $ 20 chips lost was 2 more or 2 less than the number of $ 100 chips lost , what is the largest amount of money that the gambler could ha... | "in order to maximize the amount of money that the gambler kept , we should maximize # of $ 20 chips lost and minimize # of $ 100 chips lost , which means that # of $ 20 chips lost must be 2 more than # of $ 100 chips lost . so , if # of $ 20 chips lost is x then # of $ 100 chips lost should be x - 2 . now , given that... | a ) $ 2,240 , b ) $ 2,120 , c ) $ 1,960 , d ) $ 1,920 , e ) $ 1,400 | a | subtract(multiply(const_3, const_1000), add(multiply(divide(add(14, 2), 2), 20), multiply(subtract(divide(add(14, 2), 2), 2), 100))) | add(n3,n5)|multiply(const_1000,const_3)|divide(#0,n5)|multiply(n1,#2)|subtract(#2,n5)|multiply(n2,#4)|add(#3,#5)|subtract(#1,#6)| | general |
a dress on sale in a shop is marked at $ d . during the discount sale its price is reduced by 35 % . staff are allowed a further 30 % reduction on the discounted price . if a staff member buys the dress what will she have to pay in terms of d ? | "effective discount = a + b + ab / 100 = - 35 - 30 + ( - 35 ) ( - 30 ) / 100 = - 54.5 sale price = d * ( 1 - 54.5 / 100 ) sale price = . 455 * d answer ( c )" | a ) 0.555 d , b ) 0.765 d , c ) 0.455 d , d ) 0.785 d , e ) 0.805 d | c | subtract(divide(subtract(const_100, 35), const_100), multiply(divide(subtract(const_100, 35), const_100), divide(30, const_100))) | divide(n1,const_100)|subtract(const_100,n0)|divide(#1,const_100)|multiply(#2,#0)|subtract(#2,#3)| | gain |
the number of boxes in a warehouse can be divided evenly into 6 equal shipments by boat or 28 equal shipments by truck . what is the smallest number of boxes that could be in the warehouse ? | "answer is the lcm of 6 and 28 = 84 answer c" | a ) 27 , b ) 33 , c ) 84 , d ) 81 , e ) 162 | c | multiply(multiply(multiply(multiply(const_2, const_2), const_2), const_3), 6) | multiply(const_2,const_2)|multiply(#0,const_2)|multiply(#1,const_3)|multiply(n0,#2)| | general |
if k ^ 3 is divisible by 336 , what is the least possible value of integer k ? | "k ^ 3 = 336 * x = 2 ^ 4 * 3 * 7 * x the factors of k must at minimum include 2 ^ 2 * 3 * 7 = 84 . the answer is c ." | a ) 56 , b ) 70 , c ) 84 , d ) 96 , e ) 120 | c | divide(divide(336, const_2), const_2) | divide(n1,const_2)|divide(#0,const_2)| | general |
there are 6 boxes numbered 1 , 2 , . . . 6 . each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered . the total number of ways in which this can be done is | given the constraints of this question , we have to consider 6 possible outcomes . . . 1 green ball : 6 options grrrrr rgrrrr rrgrrr rrrgrr rrrrgr rrrrrg 2 green balls : 5 options ggrrrr rggrrr rrggrr rrrggr rrrrgg 3 green balls : 4 options gggrrr rgggrr rrgggr rrrggg 4 green balls : 3 options ggggrr rggggr rrgggg 5 gr... | a ) 5 , b ) 21 , c ) 33 , d ) 60 , e ) 40 | b | add(add(add(add(6, subtract(6, 1)), subtract(subtract(6, 1), 1)), subtract(subtract(subtract(6, 1), 1), 1)), 2) | subtract(n0,n1)|add(n0,#0)|subtract(#0,n1)|add(#1,#2)|subtract(#2,n1)|add(#3,#4)|add(n2,#5) | general |
a man walked diagonally across a square lot . approximately , what was the percent saved by not walking along the edges ? | let the side of the square ( abcd ) be x metres . then , ab + bc = 2 x metres . ac = 2 x = ( 1.41 x ) m . saving on 2 x metres = ( 0.59 x ) m . saving % = ( 0.59 x ) / 2 x x 100 % = 30 % ( approx . ) b ) | ['a ) 20 %', 'b ) 30 %', 'c ) 40 %', 'd ) 50 %', 'e ) 55 %'] | b | divide(multiply(subtract(add(power(const_1, const_2), power(const_1, const_2)), sqrt(add(power(const_1, const_2), power(const_1, const_2)))), const_100), add(power(const_1, const_2), power(const_1, const_2))) | power(const_1,const_2)|add(#0,#0)|sqrt(#1)|subtract(#1,#2)|multiply(#3,const_100)|divide(#4,#1) | geometry |
rs . 500 amounts to rs . 670 in 5 years at simple interest . if the interest is increased by 2 % , it would amount to how much ? | ( 500 * 5 * 2 ) / 100 = 50 500 + 50 = 550 answer : b | a ) 150 , b ) 550 , c ) 760 , d ) 450 , e ) 260 | b | multiply(power(add(const_1, divide(2, const_100)), 5), 500) | divide(n3,const_100)|add(#0,const_1)|power(#1,n2)|multiply(n0,#2) | gain |
the lcm and hcf of two numbers are 10000 and 1000 respectively . find the larger of the two numbers if their sum is 11000 . | "there are 2 approaches in solving this . methode 1 . hcf * lcm = the actual number . 10000 * 1000 = 10000000 so the answer which we are looking for has to be a factor of 10000000 . so among the options shortlist the answers by eliminating those numbers which is not divisible by 10000000 . and then take the highest num... | a ) 10000 , b ) 562 , c ) 352 , d ) 576 , e ) 594 | a | divide(multiply(10000, 1000), 11000) | multiply(n0,n1)|divide(#0,n2)| | physics |
a cistern can be filled by a tap in 3 hours while it can be emptied by another tap in 10 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = ( 1 / 3 - 1 / 10 ) = 7 / 30 the cistern will be filled in 30 / 7 hrs i . e . , 4.3 hrs . answer : a" | a ) 4.3 hrs , b ) 8.9 hrs , c ) 2.9 hrs , d ) 7.2 hrs , e ) 8.6 hrs | a | divide(const_1, subtract(divide(const_1, 3), divide(const_1, 10))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics |
a software engineer starts from home at 3 pm for every walk on a route which has level road for some distance and a then a hillock . he walks speed of 4 kmph on level ground and then at a speed of 3 kmph on the uphill and then down the hill at a speed of 6 kmph to the level ground and then at a speed of 4 kmph to the h... | time = distance / speed so , let the distance to the level road from home be x km and the distance to the hillock from the level road be y km so , ( x / 4 + y / 3 ) + ( y / 6 + x / 4 ) = 6 x + y = 12 km . answer : a | a ) 12 km , b ) 15 km , c ) 18 km , d ) 24 km , e ) data inadequate | a | divide(multiply(subtract(9, 3), subtract(divide(add(add(6, 4), add(4, 3)), const_4), const_0_25)), const_2) | add(n1,n3)|add(n0,n1)|subtract(n5,n0)|add(#0,#1)|divide(#3,const_4)|subtract(#4,const_0_25)|multiply(#2,#5)|divide(#6,const_2) | physics |
a box contains either blue or red flags . the total number of flags in the box is an even number . a group of children are asked to pick up two flags each . if all the flags are used up in the process such that 60 % of the children have blue flags , and 60 % have red flags , what percentage of children have flags of bo... | "solution : let the total number of flags be 100 ( even number ) let the total number of ' blue ' flags alone be ' a ' let the total number of ' red ' flags alone be ' b ' let the total number of ' both ' flags be ' c ' we have given , total number of blue flags = 60 % = 60 = a + c total number of red flags = 55 % = 55... | a ) 5 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) it can not be determined | d | subtract(add(60, 60), const_100) | add(n0,n1)|subtract(#0,const_100)| | general |
there are 30 balls which are red , blue or green . if 11 balls are green and the sum of red balls and green balls is less than 25 , at most how many red balls are there ? | "r + g + b = 30 g = 11 r + g < 25 = > r + 11 < 25 = > r < 14 = > at most 13 red balls answer : b" | a ) 14 , b ) 13 , c ) 16 , d ) 17 , e ) 8 | b | subtract(11, const_1) | subtract(n1,const_1)| | general |
the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 3000 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 3000 resolutions . = 3000 * 2 * 22 / 7 * 22.4 = 422400 cm = 4224 m answer : d" | a ) 2187 m , b ) 3704 m , c ) 4179 m , d ) 4224 m , e ) 4297 m | d | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 3000), const_100) | add(const_3,const_4)|multiply(#0,const_3)|add(#1,const_1)|divide(#2,#0)|multiply(n0,#3)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,const_100)| | physics |
half a number plus 6 is 11 . what is the number ? | let x be the number . always replace ` ` is ' ' with an equal sign ( 1 / 2 ) x + 6 = 11 ( 1 / 2 ) x = 11 - 6 ( 1 / 2 ) x = 5 x = 10 correct answer is c | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | c | multiply(subtract(11, 6), const_2) | subtract(n1,n0)|multiply(#0,const_2) | general |
n ^ ( n / 2 ) = 6 is true when n = 6 in the same way what is the value of n if n ^ ( n / 2 ) = 10 ? | n ^ ( n / 2 ) = 10 apply log n / 2 logn = log 10 nlogn = 2 log 10 = log 10 ^ 2 = log 100 logn = log 100 now apply antilog n = 100 / n now n = 10 . answer : c | a ) 8 , b ) 12 , c ) 10 , d ) 16 , e ) 18 | c | divide(power(10, 2), 10) | power(n4,n0)|divide(#0,n4)| | general |
the average age of 11 boys is 50 , if the average age of first 6 boys is 49 and that of the last 6 is 52 . find the boy age ? | 1 to 11 = 11 * 50 = 550 1 to 6 = 6 * 49 = 294 6 to 11 = 6 * 52 = 312 6 th = 294 + 312 – 550 = 56 c | a ) 48 , b ) 50 , c ) 56 , d ) 62 , e ) 64 | c | subtract(multiply(6, 52), subtract(multiply(11, 50), multiply(6, 49))) | multiply(n2,n5)|multiply(n0,n1)|multiply(n2,n3)|subtract(#1,#2)|subtract(#0,#3) | general |
after successive discounts of 20 % , 10 % and 5 % a certain good is sold for rs . 6840 . find the actual price of the good . | "let actual price was 100 . after three successive discount this will become , 100 = = 20 % discount = > 80 = = 10 % discount = > 72 = = 5 % discount = 68.4 now compare , 68.4 = 6840 1 = 6840 / 68.4 100 = ( 6840 * 100 ) / 68.4 = rs . 10,000 . answer : option d" | a ) rs . 6000 , b ) rs . 9000 , c ) rs . 10800 , d ) rs . 10000 , e ) rs . 9980 | d | divide(multiply(6840, const_100), subtract(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), divide(multiply(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), 5), const_100))) | multiply(n3,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,const_100)|subtract(#1,#3)|multiply(n2,#4)|divide(#5,const_100)|subtract(#4,#6)|divide(#0,#7)| | gain |
a rope of which a calf is tied is increased from 9 m to 23 m , how much additional grassy ground shall it graze ? | "π ( 232 – 92 ) = 1408 answer : c" | a ) 1218 , b ) 1212 , c ) 1408 , d ) 2707 , e ) 2591 | c | multiply(subtract(power(23, const_2), power(const_10, const_2)), divide(add(multiply(9, const_2), const_2), add(const_4, const_3))) | add(const_3,const_4)|multiply(const_10,const_2)|power(n1,const_2)|power(n0,const_2)|add(#1,const_2)|subtract(#2,#3)|divide(#4,#0)|multiply(#6,#5)| | general |
what is the value of 3 x ^ 2 − 1.6 x + 0.3 for x = 0.8 ? | "3 x ^ 2 - 1.6 x + 0.3 for x = 0.6 = 3 ( 0.8 * 0.8 ) - 2 * 0.8 * ( 0.8 ) + 0.3 = 0.8 * 0.8 + 0.3 = 0.94 correct option : d" | a ) − 0.3 , b ) 0 , c ) 0.3 , d ) 0.94 , e ) 2.46 | d | subtract(multiply(divide(divide(subtract(power(3, 2), power(1.6, 0.3)), const_1000), const_1000), 3), divide(divide(subtract(power(3, 2), power(1.6, 0.3)), const_1000), const_1000)) | power(n0,n1)|power(n2,n3)|subtract(#0,#1)|divide(#2,const_1000)|divide(#3,const_1000)|multiply(n0,#4)|subtract(#5,#4)| | general |
in a school of 900 students , 44 % wear blue shirts , 28 % wear red shirts , 10 % wear green shirts , and the remaining students wear other colors . how many students wear other colors ( not blue , not red , not green ) ? | "44 + 28 + 10 = 82 % 100 – 82 = 18 % 900 * 18 / 100 = 162 the answer is c ." | a ) 144 , b ) 153 , c ) 162 , d ) 171 , e ) 180 | c | subtract(900, add(add(multiply(divide(44, const_100), 900), multiply(divide(28, const_100), 900)), multiply(divide(10, const_100), 900))) | divide(n1,const_100)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n0,#1)|multiply(n0,#2)|add(#3,#4)|add(#6,#5)|subtract(n0,#7)| | gain |
if c is 14 % of a and 40 % of b , what percent of a is b ? | "answer = d 14 a / 100 = 40 b / 100 b = 14 a / 40 = 35 a / 100 = 35 %" | a ) 2.5 % , b ) 15 % , c ) 25 % , d ) 35 % , e ) 250 % | d | multiply(divide(divide(14, const_100), divide(40, const_100)), const_100) | divide(n0,const_100)|divide(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)| | gain |
a new tower has just been built at the verbico military hospital ; the number of beds available for patients at the hospital is now 2 times the number available before the new tower was built . currently , 1 / 3 of the hospital ' s original beds , as well as 1 / 5 of the beds in the new tower , are occupied . for the p... | i think e - 7 / 15 is the correct answer . here goes : lets assume originally the number of beds = x after the new tower , the total combined no of beds = 2 x so old = x , new = x now 1 / 3 of x are occupied and 1 / 5 of x are occupied which simplifies to ( 1 / 5 ) x we are shifting 1 / 3 of x to the new ward so there ... | a ) 11 / 30 , b ) 29 / 60 , c ) 17 / 30 , d ) 19 / 30 , e ) 7 / 15 | e | multiply(add(multiply(2, 3), 1), multiply(divide(1, 3), divide(1, 5))) | divide(n1,n2)|divide(n1,n4)|multiply(n0,n2)|add(n1,#2)|multiply(#0,#1)|multiply(#3,#4) | general |
cricket match is conducted in us . the run rate of a cricket game was only 3.2 in first 10 over . what should be the run rate in the remaining 40 overs to reach the target of 222 runs ? | "required run rate = 222 - ( 3.2 x 10 ) = 190 = 4.75 40 40 e" | a ) 6 , b ) 6.25 , c ) 7.25 , d ) 7.5 , e ) 4.75 | e | divide(subtract(222, multiply(3.2, 10)), 40) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)| | gain |
if the sum of a number and its square is 342 , what is the number ? | "explanation : let the integer be x . then , x + x 2 = 342 x 2 + x - 342 = 0 ( x + 19 ) ( x – 18 ) = 0 x = 18 answer : e" | a ) 16 , b ) 77 , c ) 25 , d ) 87 , e ) 18 | e | floor(sqrt(342)) | sqrt(n0)|floor(#0)| | geometry |
the arithmetic mean and standard deviation of a certain normal distribution are 16.2 and 2.3 , respectively . what value is exactly 2 standard deviations less than the mean ? | "mean = 16.2 two standard deviations is 2.3 + 2.3 = 4.6 there could be two calues for this . mean + two standard deviations = 20.8 mean - two standard deviations = 11.6 answer choice has 11.6 and so c is the answer ." | a ) 10.5 , b ) 11 , c ) 11.6 , d ) 12 , e ) 12.5 | c | subtract(16.2, multiply(2, 2.3)) | multiply(n1,n2)|subtract(n0,#0)| | general |
a baker makes chocolate cookies and peanut cookies . his recipes allow him to make chocolate cookie in batches of 7 and peanut cookies in batches of 6 . if he makes exactly 99 cookies , what is the minimum number of chocolate chip cookies he makes ? | "7 c + 6 p = 99 we need to maximize p to minimize c so that the eq is also satisfied try substitution for cp to solve so that eqn is satisfied the least value of c for which equation gets satisfied is 5 i . e . 7 * 3 + 6 * 13 = 21 + 78 = 99 hence c is the answer" | a ) 7 , b ) 14 , c ) 21 , d ) 28 , e ) 35 | c | multiply(divide(subtract(99, reminder(99, add(7, 6))), add(7, 6)), 7) | add(n0,n1)|reminder(n2,#0)|subtract(n2,#1)|divide(#2,#0)|multiply(n0,#3)| | general |
p and q started a business investing rs . 85,000 and rs . 25,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | "p : q = 85000 : 25000 = 17 : 5 . answer : e" | a ) 17 : 6 , b ) 17 : 0 , c ) 17 : 4 , d ) 17 : 2 , e ) 17 : 5 | e | divide(add(multiply(add(add(2, const_3), const_3), multiply(add(2, const_3), 2)), add(2, const_3)), add(multiply(const_3, multiply(add(2, const_3), 2)), add(2, const_3))) | add(n2,const_3)|add(#0,const_3)|multiply(n2,#0)|multiply(#1,#2)|multiply(#2,const_3)|add(#0,#3)|add(#0,#4)|divide(#5,#6)| | gain |
a shopkeeper has 280 kg of apples . he sells 40 % of these at 10 % profit and remaining 60 % at 30 % profit . find his % profit on total . | if the total quantity was 100 then 40 x 10 % + 60 x 30 % = 22 this profit will remain same for any total quantity unless the % of products remains the same . hence ' a ' is the answer | a ) 22 % , b ) 25 % , c ) 26 % , d ) 28 % , e ) 35 % | a | divide(multiply(subtract(add(multiply(divide(multiply(280, 40), const_100), divide(add(const_100, 10), const_100)), multiply(divide(multiply(280, 60), const_100), divide(add(const_100, 30), const_100))), 280), const_100), 280) | add(n2,const_100)|add(n4,const_100)|multiply(n0,n1)|multiply(n0,n3)|divide(#2,const_100)|divide(#0,const_100)|divide(#3,const_100)|divide(#1,const_100)|multiply(#4,#5)|multiply(#6,#7)|add(#8,#9)|subtract(#10,n0)|multiply(#11,const_100)|divide(#12,n0) | gain |
the simple interest on a sum of money will be rs . 600 after 10 years . if the principal is trebled after 5 years , what will be the total interest at the end of the tenth year ? | solution 22.22 let the sum be rs . x . now , s . i . = rs . 600 , t = 10 years . rate = % = ( 100 x 600 / x x 10 ) % = ( 6000 / x ) % s . i . for first 5 years = rs . ( x x 5 x 6000 / x x 100 ) = rs . 300 . s . i . for first 5 years = rs . ( 3 x x 5 x 6000 / x x 100 ) = rs . 900 . ∴ total interest = rs . 1200 . answer ... | a ) rs . 600 , b ) rs . 900 , c ) 1200 rupees , d ) rs . 1500 , e ) none | c | add(multiply(multiply(divide(600, 10), 5), const_3), multiply(divide(600, 10), 5)) | divide(n0,n1)|multiply(n2,#0)|multiply(#1,const_3)|add(#2,#1) | general |
we define that k @ j is the product of j number from k in increasing order for positive integers k , j . for example , 6 @ 4 = 6 * 7 * 8 * 9 . if a = 2020 and b = 2120 , what is the value w of the ratio a / b ? | "w - > a / b = 20 * 21 * … … * 39 / 21 * 22 * … . * 39 * 40 = 20 / 40 = 1 / 2 . therefore , the answer is a ." | a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 3 , d ) 1 / 4 , e ) 1 / 5 | a | divide(divide(2020, 2020), add(divide(2020, 2020), divide(2020, 2020))) | divide(n6,n6)|add(#0,#0)|divide(#0,#1)| | general |
out of 3 consecutive odd numbers 8 times the first number is equal to addition of thrice the third number and adding 5 to twice the second . what is the first number ? | description : = > 8 x = 2 ( x + 2 ) + 5 + 3 ( x + 4 ) = > 3 x = 21 , x = 7 = > x + 4 = 11 x = 11 - 4 = 7 answer c | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | c | divide(add(add(multiply(const_2, const_2), multiply(3, const_4)), 5), subtract(8, add(3, const_2))) | add(n0,const_2)|multiply(const_2,const_2)|multiply(n0,const_4)|add(#1,#2)|subtract(n1,#0)|add(n2,#3)|divide(#5,#4) | general |
a train running at the speed of 60 km / hr crosses a pole in 30 sec . what is the length of the train ? | "speed = 60 * 5 / 18 = 50 / 3 m / sec length of the train = speed * time = 50 / 3 * 30 = 500 m answer : e" | a ) 535 m , b ) 178 m , c ) 186 m , d ) 168 m , e ) 500 m | e | multiply(divide(multiply(60, const_1000), const_3600), 30) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
a , b , c , d and e are 5 consecutive points on a straight line . if bc = 3 cd , de = 8 , ab = 5 and ac = 11 , what is the length of ae ? | "ac = 11 and ab = 5 , so bc = 6 . bc = 3 cd so cd = 2 . the length of ae is ab + bc + cd + de = 5 + 6 + 2 + 8 = 21 the answer is d ." | a ) 15 , b ) 17 , c ) 19 , d ) 21 , e ) 23 | d | add(add(11, divide(subtract(11, 5), 3)), 8) | subtract(n4,n0)|divide(#0,n1)|add(n4,#1)|add(n2,#2)| | physics |
what is the smallest number which when increased by 3 is divisible by 9 , 35 , 25 and 21 ? | "when increased by 3 , the number must include at least 3 ^ 2 * 5 ^ 2 * 7 = 1575 the answer is a ." | a ) 1572 , b ) 1782 , c ) 1992 , d ) 2122 , e ) 2342 | a | add(lcm(lcm(9, 35), lcm(25, 21)), 3) | lcm(n1,n2)|lcm(n3,n4)|lcm(#0,#1)|add(n0,#2)| | general |
if | r + s | = | r - s | , then r * s must be equal to : | solution 1 : distance perspective | r - s | = | r + s | = = > the distance of r and s is equal to the distance of r and - s . < = = = = = = = ( - s ) = = = = = = = 0 = = = = = = = ( s ) = = = = = = > only 0 is the value that has a distance equal to s and - s . solution 2 : | r - s | = | r + s | ( square both ) r ^ 2 - ... | a ) - 2 , b ) 2 , c ) 0 , d ) - 3 , e ) 3 | c | subtract(const_10, const_10) | subtract(const_10,const_10) | general |
8 cups of milk are to be poured into a 4 - cup bottle and a 8 - cup bottle . if each bottle is to be filled to the same fraction of its capacity , how many cups of milk should be poured into the 8 - cup bottle ? | let x be the # of cups going into the 8 cup bottle . so . . . . x / 8 = ( ( 8 - x ) / 4 ) 64 - 8 x = 4 x 64 = 12 x x = 16 / 3 . answer : a | a ) 16 / 3 , b ) 7 / 3 , c ) 5 / 2 , d ) 4 / 3 , e ) 3 | a | multiply(divide(8, add(4, 8)), 8) | add(n0,n1)|divide(n0,#0)|multiply(n0,#1) | general |
there are some people in party , 1 / 3 rd left the party . then 2 / 5 th of the remaining left the party , then 2 / 3 rd of the remaining left the party . at last 6 were remaining . how many people were in total ? | sol : 45 if x persons were there in total , then x × ( 1 – 1 / 3 ) × ( 1 – 2 / 5 ) × ( 1 – 2 / 3 ) = 6 x × 2 / 3 × 3 / 5 × 1 / 3 = 6 x = 6 × 5 × 3 / 2 = 45 answer : a | a ) 45 , b ) 27 , c ) 28 , d ) 26 , e ) 91 | a | divide(6, multiply(multiply(subtract(const_1, divide(1, 3)), subtract(const_1, divide(2, 5))), subtract(const_1, divide(2, 3)))) | divide(n0,n1)|divide(n2,n3)|divide(n2,n1)|subtract(const_1,#0)|subtract(const_1,#1)|subtract(const_1,#2)|multiply(#3,#4)|multiply(#6,#5)|divide(n6,#7) | general |
in what ratio must tea at rs . 62 per kg be mixed with tea at rs . 72 per kg so that the mixture must be worth rs . 67 per kg ? | "required ratio = 500 : 500 = 3 : 1 answer e" | a ) 3 : 1 , b ) 3 : 2 , c ) 4 : 3 , d ) 5 : 3 , e ) 1 : 1 | e | divide(subtract(72, 67), subtract(67, 62)) | subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1)| | other |
a 8 % stock yielding 10 % is quoted at : | "solution to earn rs . 10 , money invested = rs . 100 . to earn rs . 8 , money invested = rs . ( 100 / 10 x 8 ) = rs . 80 . â ˆ ´ market value of rs . 100 stock = rs . 80 answer a" | a ) 80 , b ) 110 , c ) 112 , d ) 120 , e ) 160 | a | multiply(divide(const_100, 10), 8) | divide(const_100,n1)|multiply(n0,#0)| | gain |
if it is assumed that 50 percent of those who receive a questionnaire by mail will respond and 300 responses are needed , what is the minimum number of questionnaires that should be mailed ? | "minimum no of mail to be sent for getting 300 responses at 50 % = 300 / 0.5 = 600 option e" | a ) 400 , b ) 420 , c ) 480 , d ) 500 , e ) 600 | e | divide(300, divide(50, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain |
an error 8 % in excess is made while measuring the side of a square . now what is the percentage of error in the calculated area of the square ? | "percentage error in calculated area = ( 8 + 8 + ( 8 × 8 ) / 100 ) % = 16.64 % answer : b" | a ) 6.64 % , b ) 16.64 % , c ) 15.64 % , d ) 26.64 % , e ) 10.64 % | b | divide(multiply(subtract(square_area(add(const_100, 8)), square_area(const_100)), const_100), square_area(const_100)) | add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)| | gain |
the average age of 15 students of a class is 15 years . out of these , the average age of 7 students is 14 years and that of the other 7 students is 16 years . the age of the 15 th student is | "solution age of the 15 th student = [ 15 x 15 - ( 14 x 7 + 16 x 7 ) ] = ( 225 - 210 ) = 15 years . answer c" | a ) 9 years , b ) 11 years , c ) 15 years , d ) 21 years , e ) 25 years | c | subtract(multiply(15, 15), add(multiply(7, 14), multiply(7, 16))) | multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)| | general |
car a travels at 3 times the average speed of car b . car a started to travel at 12 : 00 o ' clock , car b started to travel at 16 : 00 o ' clock . what is the speed of car b ( in km / h ) if the total distance that both cars traveled until 18 : 00 was 1000 km ? | total distance - 1000 km time ratio a : b - 3 : 1 speed ratio a : b - 3 : 1 so distance ( speed * time ) ratio will be - 3 * 3 : 1 * 1 - 9 : 1 so total distance traveled by b - 100 km time taken by b ( 18 : 00 - 16 : 00 = 2 ) - 2 hours speed of b - 50 kmph answer : e | a ) 10 . , b ) 25 . , c ) 30 . , d ) 38 , e ) 50 . | e | divide(1000, add(multiply(subtract(18, 12), 3), subtract(18, 16))) | subtract(n5,n1)|subtract(n5,n3)|multiply(n0,#0)|add(#2,#1)|divide(n7,#3) | general |
there are 6 books on a shelf , of which 2 are paperbacks and 4 are hardbacks . how many possible selections of 4 books from this shelf include at least one paperback ? | "approach 1 at - least 1 paper back = total - no paper back 6 c 4 - 4 c 4 = 14 approach 2 at - least 1 paper back = 1 paper back , 3 hard back or 2 paper back 2 hard back = 2 c 1 * 4 c 3 + 2 c 2 * 4 c 2 = 14 answer is b" | a ) a ) 40 , b ) b ) 14 , c ) c ) 50 , d ) d ) 55 , e ) e ) 60 | b | subtract(choose(6, 4), choose(4, 4)) | choose(n0,n3)|choose(n2,n3)|subtract(#0,#1)| | probability |
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 . 931 . 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 ) = 931 96 + 225 + 490 + 6 x = 931 6 x = 120 = > x = 20 . answer : e" | a ) 25 , b ) 76 , c ) 29 , d ) 12 , e ) 20 | e | divide(subtract(subtract(subtract(931, 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 |
the average weight of 10 men is increased by 2 ½ kg when one of the men who weighs 68 kg is replaced by a new man . what is the weight of the new man ? | "since the average has increased by 1.5 kg , the weight of the man who stepped in must be equal to 68 + 10 x 2.5 68 + 25 = 93 kg ans : ' b '" | a ) 80 kg , b ) 93 kg , c ) 70 kg , d ) 75 kg , e ) 85 kg | b | add(68, multiply(10, add(2, divide(const_1, 2)))) | divide(const_1,n1)|add(n1,#0)|multiply(n0,#1)|add(n2,#2)| | general |
in a certain company , a third of the workers do not have a retirement plan . 20 % of the workers who do not have a retirement plan are women , and 40 % of the workers who do have a retirement plan are men . if 176 of the workers of that company are men , how many of the workers are women ? | "set up equation : x = total number of workers 176 = 0,4 * 2 / 3 * x + 0,8 * 1 / 3 * x 176 = 16 / 30 x x = 330 330 - 176 = 154 answer e" | a ) 80 , b ) 95 , c ) 105 , d ) 120 , e ) 154 | e | multiply(divide(176, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(20, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))), add(multiply(divide(const_1, const_3), divide(20, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), mul... | divide(const_1,const_3)|divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|subtract(const_1,#0)|multiply(#2,#4)|subtract(#0,#3)|add(#5,#6)|subtract(#4,#5)|add(#3,#8)|divide(n2,#7)|multiply(#9,#10)| | gain |
elena ’ s bread recipe calls for 3 ounces of butter for each 4 cups of flour used . she needs to make 6 times the original recipe . if 12 ounces of butter is used , then how many cups of flour are needed ? | "solving through algebra route : 3 b + 4 f = x amount if we multiply this equation with 6 we get : 18 b + 24 f = 6 x therefore , we got 18 ounces of butter and 6 x amount of quantity when we use 24 ounces of floor . ans : e" | a ) 1 , b ) 4 , c ) 9 , d ) 13 , e ) 24 | e | multiply(4, 6) | multiply(n1,n2)| | general |
in how many w ways can 5 people from a group of 6 people be seated around a circular table | w = 6 c 5 * ( 5 - 1 ) ! ( select 5 out of 6 and arrange them in circular manner ) = 6 * 4 ! = 6 * 24 = 144 answer - e | ['a ) 56', 'b ) 80', 'c ) 100', 'd ) 120', 'e ) 144'] | e | divide(factorial(6), 5) | factorial(n1)|divide(#0,n0) | geometry |
suganya and suriya are partners in a business . suganya invests rs . 35,000 for 8 months and suriya invests rs . 42,000 for 10 months . out of a profit of rs . 30,570 . suganya ' s share is | solution ratio of their shares = ( 35000 ã — 8 ) : ( 42000 ã — 10 ) = 128 : 125 suganya ' s share = rs . ( 30570 ã — 2 / 253 ) = rs . 241.7 . answer d | a ) rs . 9471 , b ) rs . 12,628 , c ) rs . 262.8 , d ) rs . 241.7 , e ) none | d | multiply(multiply(const_0_25, const_100), 10) | multiply(const_0_25,const_100)|multiply(n3,#0) | gain |
a space shuttle orbits the earth at about 9 kilometers per second . this speed is equal to how many kilometers per hour ? | "seconds in 1 hours : 60 s in 1 min 60 min in 1 hr 60 * 60 = 3600 sec in 1 hr 9 * 3600 = 32,400 answer : e" | a ) 480 , b ) 2,880 , c ) 4,800 , d ) 28,800 , e ) 32,400 | e | multiply(9, const_3600) | multiply(n0,const_3600)| | physics |
7 liters of a 22 percent solution of alcohol in water are mixed with 3 liters of an 11 percent alcohol in water solution . what is the percentage of alcohol in the new solution ? | "the total amount of alcohol is 0.22 ( 7 ) + 0.11 ( 3 ) = 1.87 liters . the percentage is 1.87 / 10 = 18.7 / 100 which is 18.7 % the answer is e ." | a ) 17.5 % , b ) 17.8 % , c ) 18.1 % , d ) 18.4 % , e ) 18.7 % | e | multiply(divide(add(multiply(7, divide(22, const_100)), multiply(divide(11, const_100), multiply(7, divide(22, const_100)))), add(7, const_2.0)), const_100) | add(n0,const_2.0)|divide(n1,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(#2,#3)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)| | general |
a cricketer scored 136 runs which included 12 boundaries and 2 sixes . what percent of his total score did he make by running between the wickets . | "explanation : number of runs made by running = 136 - ( 12 x 4 + 2 x 6 ) = 136 - ( 60 ) = 76 now , we need to calculate 72 is what percent of 136 . = > 76 / 136 * 100 = 55.88 % answer : c" | a ) 46.2 % , b ) 54.54 % , c ) 55.88 % , d ) 70 % , e ) none of these | c | multiply(divide(subtract(136, add(multiply(12, const_4), multiply(2, multiply(2, const_3)))), 136), const_100) | multiply(n1,const_4)|multiply(n2,const_3)|multiply(n2,#1)|add(#0,#2)|subtract(n0,#3)|divide(#4,n0)|multiply(#5,const_100)| | general |
by investing in 1623 % stock at 64 , one earns rs . 1700 . the investment made is | "explanation : market value = rs . 64 face value is not given and hence take it as rs . 100 16 2 / 3 % of the face value = 50 / 3 ie , to earn 50 / 3 , investment = rs . 64 hence , to earn rs . 1700 , investment needed = 64 × 3 × 1700 / 50 = 6528 answer : option d" | a ) s . 9600 , b ) s . 7500 , c ) s . 5640 , d ) s . 6528 , e ) s . 6760 | d | multiply(divide(multiply(multiply(multiply(const_4, const_2), multiply(const_4, const_2)), const_3), divide(const_100, const_2)), subtract(1700, multiply(const_4, const_100))) | divide(const_100,const_2)|multiply(const_2,const_4)|multiply(const_100,const_4)|multiply(#1,#1)|subtract(n2,#2)|multiply(#3,const_3)|divide(#5,#0)|multiply(#6,#4)| | gain |
for any integer k greater than 1 , the symbol k * denotes the product of all integers between 1 and k , inclusive . if k * is a multiple of 945 what is the least possible value of k ? | "945 = 3 * 3 * 3 * 5 * 7 thus k must include numbers at least up to the number 9 so that there are at least three appearances of 3 ( that is : 3 , 6 , and 9 ) . the answer is c ." | a ) 5 , b ) 7 , c ) 9 , d ) 11 , e ) 13 | c | divide(divide(divide(945, const_3), const_3), add(1, const_4)) | add(n0,const_4)|divide(n2,const_3)|divide(#1,const_3)|divide(#2,#0)| | general |
if ( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 2 ) ( 3 ^ b ) and a and b are positive integers , what is the value of a ? | "( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 2 ) ( 3 ^ b ) = 2 ^ a . 9 ^ a . 9 ^ ( 3 a – 1 ) = ( 2 ^ 2 ) ( 3 ^ b ) just compare powers of 2 from both sides answer = 2 = e" | a ) 22 , b ) 11 , c ) 9 , d ) 6 , e ) 2 | e | multiply(3, 1) | multiply(n2,n3)| | general |
there are 6 people in the elevator . their average weight is 160 lbs . another person enters the elevator , and increases the average weight to 151 lbs . what is the weight of the 7 th person . | "solution average of 7 people after the last one enters = 151 . â ˆ ´ required weight = ( 7 x 151 ) - ( 6 x 160 ) = 1057 - 960 = 97 . answer a" | a ) 97 , b ) 168 , c ) 189 , d ) 190 , e ) 200 | a | subtract(multiply(151, 7), multiply(6, 160)) | multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)| | general |
what least fraction should be added to 1 / ( 2 × 3 ) + 1 / ( 3 × 4 ) + 1 / ( 4 × 5 ) + . . . . . . . + 1 / ( 21 × 22 ) so that the result is unity ? | explanation : given expression = 1 / ( 2 × 3 ) + 1 / ( 3 × 4 ) + 1 / ( 4 × 5 ) + . . . . . . + 1 / ( 21 × 22 ) = ( 1 / 2 - 1 / 3 ) + ( 1 / 3 - 1 / 4 ) + ( 1 / 4 - 1 / 5 ) + . . . . . . . . . . = ( 1 / 21 - 1 / 22 ) = 1 / 2 - 1 / 22 = 10 / 22 = 5 / 11 thus , least fraction to be added = 1 - 5 / 11 = 6 / 11 answer : opti... | a ) 5 / 11 , b ) 6 / 11 , c ) 4 / 11 , d ) 7 / 11 , e ) 8 / 11 | b | divide(multiply(3, 2), add(multiply(3, 2), add(2, 3))) | add(n1,n2)|multiply(n1,n2)|add(#0,#1)|divide(#1,#2) | general |
how many positive integers less than 160 are there such that they are multiples of 13 or multiples of 12 but not both ? | "for 13 : 13 . . . 160 = 13 * 12 for 12 : 12 . . . 156 = 12 * 13 but there is one integer 13 * 12 . so n = ( 12 - 1 ) + ( 13 - 1 ) = 23 d" | a ) 18 , b ) 19 , c ) 20 , d ) 23 , e ) 25 | d | divide(factorial(subtract(add(const_4, 13), const_1)), multiply(factorial(13), 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 |
24 * 22 = ? / 8 | "a 4224 ? / 8 = 24 * 22 ? = 24 * 22 * 8 ? = 4224" | a ) 4224 , b ) 2224 , c ) 5374 , d ) 2434 , e ) 4254 | a | multiply(24, 22) | multiply(n0,n1)| | general |
how many 5 - digit numbers that do not contain the digits 4 or 7 are there ? | we can have 7 digits ( 1 , 2,3 , 5,6 , 8,9 ) for the first place ( ten thousand ' s place ) . and similarly 8 digits for thousand ' s , hundred ' s , tenth ' s and unit digit . ( 0,1 , 2,3 , 5,6 , 8,9 ) so in total 7 * 8 * 8 * 8 * 8 = 28672 hence c | a ) 44648 , b ) 27844 , c ) 28642 , d ) 16864 , e ) 32458 | c | multiply(power(multiply(const_4, const_2), subtract(5, const_1)), 7) | multiply(const_2,const_4)|subtract(n0,const_1)|power(#0,#1)|multiply(n2,#2) | general |
evaluate 75 % of 450 + 45 % of 750 | "explanation : = ( 75 / 100 ) * 450 + ( 45 / 100 ) * 750 = 675 option e" | a ) 632 , b ) 642 , c ) 652 , d ) 675 , e ) 572 | e | divide(75, divide(450, 75)) | divide(n1,n0)|divide(n0,#0)| | gain |
how many positive integers i less than 20 are either a multiple of 2 , an odd multiple of 9 , or the sum of a positive multiple of 2 and a positive multiple of 9 ? | we ' re asked to deal with the positive integers less than 20 . there are only 19 numbers in that group ( 1 to 19 , inclusive ) . we ' re asked to find all of the numbers that fit one ( or more ) of the given descriptions . looking at the answer choices , we have every value form 15 to 19 , inclusive , so most ( if not... | a ) 19 , b ) 18 , c ) 17 , d ) 16 , e ) 15 | e | add(add(9, const_1), add(2, const_3)) | add(n2,const_1)|add(n1,const_3)|add(#0,#1) | general |
a train passes a station platform in 33 sec and a man standing on the platform in 20 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 20 = 300 m . let the length of the platform be x m . then , ( x + 300 ) / 33 = 15 = > x = 195 m . answer : d" | a ) 228 , b ) 240 , c ) 887 , d ) 195 , e ) 1811 | d | multiply(20, multiply(54, const_0_2778)) | multiply(n2,const_0_2778)|multiply(n1,#0)| | physics |
the weights of 3 bags are in the ratio 4 : 5 : 6 . if the sum of the weights of the heaviest and the lightest bag is 45 kg more than the weight of the third bag , what is the weight of the lightest bag ? | let the weights of the three boys be 4 k , 5 k and 6 k respectively . 4 k + 6 k = 5 k + 45 = > 5 k = 45 = > k = 9 therefore the weight of the lightest boy = 4 k = 4 ( 9 ) = 36 kg . answer : d | a ) 45 kg , b ) 38 kg , c ) 66 kg , d ) 36 kg , e ) of these | d | multiply(divide(45, subtract(add(4, 6), 5)), 4) | add(n1,n3)|subtract(#0,n2)|divide(n4,#1)|multiply(n1,#2) | general |
the area of a circular field is 13.86 hectares . find the cost of fencing it at the rate of rs . 4.70 per metre . | "explanation : area = ( 13.86 x 10000 ) sq . m = 138600 sq . m circumference = cost of fencing = rs . ( 1320 x 4.70 ) = rs . 6204 . answer : b ) 6204" | a ) 2399 , b ) 6204 , c ) 2999 , d ) 5808 , e ) 2888 | b | multiply(circumface(multiply(sqrt(divide(13.86, const_pi)), const_100)), 4.70) | divide(n0,const_pi)|sqrt(#0)|multiply(#1,const_100)|circumface(#2)|multiply(#3,n1)| | geometry |
if x / 4 + 24 / x = 5 , what are the values of 4 x - 7 ? | i got the same thing c is the answer 25 or 41 | a ) 8 and 9 , b ) 8 and 24 , c ) 25 and 41 , d ) 12 and 29 , e ) 17 and 29 | c | add(multiply(subtract(add(subtract(24, 4), sqrt(subtract(power(subtract(24, 4), 4), multiply(4, multiply(24, 4))))), 7), const_10), subtract(subtract(subtract(24, 4), sqrt(subtract(power(subtract(24, 4), 4), multiply(4, multiply(24, 4))))), 7)) | multiply(n0,n1)|subtract(n1,n0)|multiply(n0,#0)|power(#1,n3)|subtract(#3,#2)|sqrt(#4)|add(#5,#1)|subtract(#1,#5)|subtract(#6,n4)|subtract(#7,n4)|multiply(#8,const_10)|add(#10,#9)| | general |
a room is 6 meters 72 centimeters in length and 4 meters 32 centimeters in width . find the least number of square tiles of equal size required to cover the entire floor of the room . | let us calculate both the length and width of the room in centimeters . length = 6 meters and 72 centimeters = 672 cm width = 4 meters and 32 centimeters = 432 cm as we want the least number of square tiles required , it means the length of each square tile should be as large as possible . further , the length of each ... | a ) 107 , b ) 126 , c ) 117 , d ) 136 , e ) 147 | b | divide(multiply(add(multiply(6, const_100), 72), add(multiply(4, const_100), 32)), multiply(gcd(add(multiply(6, const_100), 72), add(multiply(4, const_100), 32)), gcd(add(multiply(6, const_100), 72), add(multiply(4, const_100), 32)))) | multiply(n0,const_100)|multiply(n2,const_100)|add(n1,#0)|add(n3,#1)|gcd(#2,#3)|multiply(#2,#3)|multiply(#4,#4)|divide(#5,#6) | general |
a bottle contains a certain solution . in the bottled solution , the ratio of water to soap is 3 : 4 , and the ratio of soap to salt is four times this ratio . the solution is poured into an open container , and after some time , the ratio of water to soap in the open container is halved by water evaporation . at that ... | "water : soap = 3 : 4 soap : salt = 12 : 16 = > for 12 soap , salt = 16 = > for 4 soap , salt = ( 16 / 12 ) * 4 = 16 / 12 = 64 / 12 = 18 / 3 = 6 / 1 so , water : soap : salt = 3 : 4 : 6 = 3 : 4 : 6 after open container , water : soap : salt = 1.5 : 4 : 6 so , water : salt = 1.5 : 6 = 3 : 12 answer : d" | a ) 3 : 8 , b ) 3 : 9 , c ) 3 : 4 , d ) 3 : 12 , e ) 4 : 12 | d | divide(multiply(multiply(4, const_3.0), 3), multiply(multiply(4, 4), 4)) | multiply(n0,const_2.0)|multiply(n1,n1)|multiply(n0,#0)|multiply(n1,#1)|divide(#2,#3)| | other |
in a group of people , if 30 people were made to stand in each column , 16 columns could be formed . if 12 people were made to stand in a column , how many columns could be formed ? | "16 * 30 = 12 * n n = 40 the answer is c ." | a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | c | divide(30, divide(12, 16)) | divide(n2,n1)|divide(n0,#0)| | other |
if x is a positive integer with fewer than 3 digits , what is the probability q that x * ( x + 1 ) is a multiple of either 4 or 5 ? | interesting question ! also one that we should be able to answer very quickly be keeping an eye on our best friends , the answer choices . we know that x belongs to the set { 1 , 2 , 3 , . . . , 99 } . we want to know the probability q that x ( x + 1 ) is a multiple of either 4 or 5 . when will this happen ? if either ... | a ) 4 / 99 , b ) 2 / 25 , c ) 8 / 99 , d ) 49 / 100 , e ) 86 / 99 | e | divide(subtract(add(multiply(divide(const_100, 4), const_2), multiply(divide(const_100, 5), const_2)), 4), subtract(const_100, const_1)) | divide(const_100,n2)|divide(const_100,n3)|subtract(const_100,const_1)|multiply(#0,const_2)|multiply(#1,const_2)|add(#3,#4)|subtract(#5,n2)|divide(#6,#2) | general |
by weight , liquid x makes up 0.8 percent of solution a and 1.8 percent of solution b . if 200 grams of solution a are mixed with 700 grams of solution b , then liquid x accounts for what percent of the weight of the resulting solution ? | "i think there is a typo in question . it should have been ` ` by weight liquid ' x ' makes up . . . . . ` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 200 gms of a and 700 gms of b is mixed : weight of liquid x = ( 0.8 * 200 ) / 100 + ( 1.8 * 700 ) / 100 = 14.2 gms % of liquid x in resultant... | a ) 1.42 % , b ) 1.9 % , c ) 10 % , d ) 15 % , e ) 19 % | a | divide(add(multiply(200, 0.8), multiply(700, 1.8)), const_1000) | multiply(n0,n2)|multiply(n1,n3)|add(#0,#1)|divide(#2,const_1000)| | gain |
the s . i . on a certain sum of money for 3 years at 5 % per annum is half the c . i . on rs . 4000 for 2 years at 10 % per annum . the sum placed on s . i . is ? | "explanation : c . i . = [ 4000 * ( 1 + 10 / 100 ) 2 - 4000 ] = ( 4000 * 11 / 10 * 11 / 10 - 4000 ) = rs . 840 . sum = ( 420 * 100 ) / ( 3 * 5 ) = rs . 2800 answer : b" | a ) 2197 , b ) 2800 , c ) 1750 , d ) 2267 , e ) 1262 | b | divide(divide(subtract(multiply(4000, power(add(const_1, divide(10, const_100)), 2)), 4000), 2), multiply(3, divide(5, const_100))) | divide(n4,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|power(#2,n3)|multiply(n2,#4)|subtract(#5,n2)|divide(#6,n3)|divide(#7,#3)| | gain |
two train each 500 m long , are running in opposite directions on parallel tracks . if their speeds are 45 km / hr and 30 km / hr respectively , the time taken by the slower train to pass the driver of the faster one is | "explanation : relative speed = 45 + 30 = 75 km / hr = 750 / 36 m / s = 125 / 6 m / s we are calculating the time taken by the slower train to pass the driver of the faster one . hence the distance = length of the smaller train = 500 m time = distance / speed = 500 / ( 125 / 6 ) = 24 sec answer : option c" | a ) 50 sec , b ) 58 sec , c ) 24 sec , d ) 22 sec , e ) 21 sec | c | divide(500, multiply(add(45, 30), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
if a fraction is multiplied by itself and then divided by the reciprocal of the same fraction , the result is 18 26 / 27 . find the fraction . | answer if the required fraction be p according to the question ( p x p ) / ( 1 / p ) = 1826 / 27 ⇒ p 3 = 512 / 27 ∴ p = 8 / 3 = 2 2 / 3 correct option : c | a ) 8 / 27 , b ) 1 1 / 3 , c ) 2 2 / 3 , d ) 3 / 3 , e ) none of these | c | add(subtract(multiply(power(add(divide(26, 27), 18), divide(const_1, const_3)), const_3), power(add(divide(26, 27), 18), divide(const_1, const_3))), const_2) | divide(n1,n2)|divide(const_1,const_3)|add(n0,#0)|power(#2,#1)|multiply(#3,const_3)|subtract(#4,#3)|add(#5,const_2) | general |
two trains of length 100 meters and 200 meters are 630 meters apart . they are moving towards each other on parallel tracks , at speeds of 90 km / h and 72 km / h . after how many seconds will the trains meet ? | the speeds are 90000 / 3600 = 25 m / s and 72000 / 3600 = 20 m / s the relative speed is 45 m / s . time = 630 / 45 = 14 seconds the answer is b . | a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | b | divide(630, multiply(add(90, 72), const_0_2778)) | add(n3,n4)|multiply(#0,const_0_2778)|divide(n2,#1) | physics |
a can do a job in 15 days and b can do it in 30 days . a and b working together will finish twice the amount of work in days ? | "1 / 15 + 1 / 30 = 3 / 30 = 1 / 10 10 = 10 * 2 = 20 days answer : a" | a ) 20 , b ) 22 1 / 2 , c ) 22 1 / 0 , d ) 22 1 / 9 , e ) 22 1 / 1 | a | add(divide(const_1, 15), divide(const_1, 30)) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)| | physics |
find the value of y from ( 12 ) ^ 3 x 6 ^ 4 ÷ 432 = y ? | "given exp . = ( 12 ) 3 x 64 = ( 12 ) 3 x 64 = ( 12 ) 2 x 62 = ( 72 ) 2 = 5184 432 12 x 62 e" | a ) 2450 , b ) 2560 , c ) 2650 , d ) 3500 , e ) 5184 | e | divide(multiply(power(12, 3), power(6, 4)), 432) | power(n0,n1)|power(n2,n3)|multiply(#0,#1)|divide(#2,n4)| | general |
in a graduating class , 40 percent of the students are male . in this class , 60 percent of the male students and 40 percent of the female students are 25 years old or older . if one student in the class is randomly selected , approximately what is the probability that he or she will be less than 25 years old ? | "let x be the total number of students . the number students who are younger than 25 is 0.4 * 0.4 x + 0.6 * 0.6 x = 0.52 x the answer is d ." | a ) 0.4 , b ) 0.44 , c ) 0.48 , d ) 0.52 , e ) 0.56 | d | subtract(const_1, add(multiply(divide(40, const_100), divide(60, const_100)), multiply(divide(40, const_100), subtract(const_1, divide(40, const_100))))) | divide(n0,const_100)|divide(n1,const_100)|divide(n2,const_100)|multiply(#0,#1)|subtract(const_1,#0)|multiply(#2,#4)|add(#3,#5)|subtract(const_1,#6)| | general |
the length of the bridge , which a train 130 metres long and travelling at 45 km / hr can cross in 30 seconds , is ? | "speed = [ 45 x 5 / 18 ] m / sec = [ 25 / 2 ] m / sec time = 30 sec let the length of bridge be x metres . then , ( 130 + x ) / 30 = 25 / 2 = > 2 ( 130 + x ) = 750 = > x = 245 m . answer : c" | a ) 245 , b ) 77 , c ) 88 , d ) 99 , e ) 22 | c | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 130) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
at a tanning salon , customers are charged $ 10 for their first visit in a calendar month and $ 8 for each visit after that in the same calendar month . in the last calendar month , 100 customers visited the salon , of which 30 made a second visit , and 10 made a third visit . all other customers made only one visit . ... | i get b . this question seems too straightforward for 600 + . am i missing something ? 100 first - time visits - - > 100 ( 10 ) = $ 1000 30 + 10 = 40 subsequent visits - - > 40 ( 8 ) = $ 320 total revenue : 1000 + 320 = $ 1320 the answer is b . | a ) $ 1220 , b ) $ 1240 , c ) $ 1300 , d ) $ 1340 , e ) $ 1880 | b | add(multiply(add(10, 8), 30), multiply(subtract(100, 30), 10)) | add(n0,n1)|subtract(n2,n3)|multiply(n3,#0)|multiply(n0,#1)|add(#2,#3) | physics |
if n is a positive integer , what is the remainder when ( 7 ^ ( 4 n + 2 ) ) ( 6 ^ n ) is divided by 10 ? | this one took me bout 3 1 / 2 min . just testin numbers and what not . first notice that n is positive . save time by noticing thati worked out one solution where n = 0 only to find that thats not an option : p . 1 - 7 stands for ^ 1 thru 7 1 : 7 * 1 = 7 2 : 7 * 7 = 9 3 : 7 * 9 = 3 4 : 7 * 3 = 1 5 : 7 * 1 = 7 6 : 7 * 7... | a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | c | reminder(multiply(multiply(const_3, const_3), 6), 10) | multiply(const_3,const_3)|multiply(n3,#0)|reminder(#1,n4) | general |
audrey 4 hours to complete a certain job . ferris can do the same job in 3 hours . audrey and ferris decided to collaborate on the job , working at their respective rates . while audrey worked continuously , ferris took 1 breaks of equal length . if the two completed the job together in 2 hours , how many minutes long ... | "audery and ferris collective work rate : 1 / 4 + 1 / 3 = 7 / 12 collective work time = 12 / 7 = 1.7 hrs job was actually done in = 2 ( includes breaks ) breaks = actual time taken - collective work time = 2 - 1.7 = . 3 hrs = 1 / 2 so ferrais took 1 breaks = . 3 / 1 = . 3 hrs = 30 m so answer is b ) 30 mins" | a ) 5 , b ) 30 , c ) 15 , d ) 20 , e ) 25 | b | divide(subtract(multiply(2, multiply(1, const_10)), multiply(divide(3, 2), multiply(1, const_10))), 1) | divide(n1,n3)|multiply(n2,const_10)|multiply(n3,#1)|multiply(#0,#1)|subtract(#2,#3)|divide(#4,n2)| | physics |
sahil purchased a machine at rs 11000 , then got it repaired at rs 5000 , then gave its transportation charges rs 1000 . then he sold it with 50 % of profit . at what price he actually sold it . | "explanation : question seems a bit tricky , but it is very simple . just calculate all cost price , then get 150 % of cp . c . p . = 11000 + 5000 + 1000 = 17000 150 % of 17000 = 150 / 100 * 17000 = 25500 option b" | a ) rs . 22500 , b ) rs . 25500 , c ) rs . 26000 , d ) rs . 28000 , e ) none of these | b | add(add(add(11000, 5000), 1000), multiply(divide(add(add(11000, 5000), 1000), const_100), 50)) | add(n0,n1)|add(n2,#0)|divide(#1,const_100)|multiply(n3,#2)|add(#1,#3)| | gain |
a boat can travel with a speed of 16 km / hr in still water . if the speed of the stream is 4 km / hr , find the time taken by the boat to go 60 km downstream . | "speed downstream = ( 16 + 4 ) km / hr = 20 km / hr . time taken to travel 60 km downstream = 60 / 20 = 3 hours . answer : b" | a ) 2 hours . , b ) 3 hours . , c ) 4 hours . , d ) 5 hours . , e ) 6 hours . | b | divide(60, add(16, 4)) | add(n0,n1)|divide(n2,#0)| | physics |
walking 5 / 4 of his usual rate , a boy reaches his school 4 min early . find his usual time to reach the school ? | "speed ratio = 1 : 5 / 4 = 4 : 5 time ratio = 5 : 4 1 - - - - - - - - 5 4 - - - - - - - - - ? 20 m . answer : e" | a ) 22 , b ) 99 , c ) 27 , d ) 28 , e ) 20 | e | multiply(4, 5) | multiply(n0,n2)| | gain |
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 46.5 meters . how many times did the ball bounce ? | "going down = 16 m going up = 8 - - > total = 24 going down = 8 - - > total = 32 going up = 4 - - > total = 36 going down = 4 - - > total = 40 going up = 2 - - > total = 42 going down = 2 - - > total = 44 going up = 1 - - > total = 45 going down = 1 - - > total = 46 going up = 0.5 - - > total = 46.5 ( caught ) no of go... | a ) 5 , 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 |
6000 - 5000 ÷ 20.00 = ? | "answer given expression = 6000 - 5000 ÷ 20.00 = 6000 - 250 = 5750 correct option : c" | a ) 900 , b ) 500 , c ) 5750 , d ) 5000 , e ) none | c | subtract(multiply(divide(6000, const_100), 5000), multiply(divide(const_1, const_3), multiply(divide(6000, const_100), 5000))) | divide(n0,const_100)|divide(const_1,const_3)|multiply(n1,#0)|multiply(#1,#2)|subtract(#2,#3)| | general |
mahesh can do a piece of work in 50 days . he works at it for 20 days and then rajesh finished it in 30 days . how long will y take to complete the work ? | "work done by mahesh in 50 days = 20 * 1 / 50 = 2 / 5 remaining work = 1 - 2 / 5 = 3 / 5 3 / 5 work is done by rajesh in 30 days whole work will be done by rajesh is 30 * 5 / 3 = 50 days answer is e" | a ) 45 , b ) 25 , c ) 37 , d ) 41 , e ) 50 | e | divide(const_1, divide(subtract(const_1, multiply(20, divide(const_1, 50))), 30)) | divide(const_1,n0)|multiply(n1,#0)|subtract(const_1,#1)|divide(#2,n2)|divide(const_1,#3)| | physics |
the area of a square is 4096 sq cm . find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is 24 cm less than the side of the square . | let the length and the breadth of the rectangle be l cm and b cm respectively . let the side of the square be a cm . a 2 = 4096 = 212 a = ( 212 ) 1 / 2 = 26 = 64 l = 2 a and b = a - 24 b : l = a - 24 : 2 a = 40 : 128 = 5 : 16 answer : e | ['a ) 5 : 78', 'b ) 5 : 11', 'c ) 5 : 12', 'd ) 5 : 19', 'e ) 5 : 16'] | e | divide(subtract(sqrt(4096), 24), multiply(sqrt(4096), const_2)) | sqrt(n0)|multiply(#0,const_2)|subtract(#0,n1)|divide(#2,#1) | geometry |
a baker makes chocolate cookies and peanut cookies . his recipes allow him to make chocolate cookie in batches of 5 and peanut cookies in batches of 6 . if he makes exactly 94 cookies , what is the minimum number of chocolate chip cookies he makes ? | 5 c + 6 p = 94 we need to maximize p to minimize c so that the eq is also satisfied try substitution for cp to solve so that eqn is satisfied the least value of c for which equation gets satisfied is 5 i . e . 5 * 8 + 6 * 9 = 40 + 54 = 94 hence e is the answer | a ) 7 , b ) 14 , c ) 21 , d ) 28 , e ) 40 | e | multiply(divide(subtract(94, reminder(94, add(5, 6))), add(5, 6)), 5) | add(n0,n1)|reminder(n2,#0)|subtract(n2,#1)|divide(#2,#0)|multiply(n0,#3) | general |
when y is divided by 264 , the remainder is 42 . what is the remainder when the same y is divided by 22 ? | "y = 264 * a + 42 = ( 22 * 12 ) * a + 22 + 20 the answer is e ." | a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | e | reminder(42, 22) | reminder(n1,n2)| | general |
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