options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) 5 , b ) 305 , c ) 365 , d ) 50 , e ) 495 | d | add(add(4, 7), 7) | how many 4 - digit even numbers are possible such that if one of the digits is 5 , the next / succeeding digit to it should be 7 ? | 5700 , 5710 , . . . . . . 5798 , so total 50 . hence option d . | a = 4 + 7
b = a + 7
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a ) 116 , b ) 120 , c ) 130 , d ) 140 , e ) none | c | add(divide(1044, 9), floor(divide(divide(1044, 9), 9))) | a man has 1044 candles . after burning , he can make a new candle from 9 stubs left behind . find the maximum number of candles that can be made . | explanation : since , 1 candle is made from 9 stubs . therefore , from 1044 stubs , 1044 / 9 i . e 116 candles will be made . now , from 116 stubs , 116 / 9 i . e 12 candles can be made and 8 stubs will be left . thus , total stubs left is ( 12 + 8 ) = 20 out of which 20 / 9 i . e 2 candles can be made and 2 stubs will be left . thus , maximum number of candles that can be made is 130 ( 116 + 12 + 2 ) . answer : c | a = 1044 / 9
b = 1044 / 9
c = b / 9
d = math.floor(c)
e = a + d
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a ) 5568 , b ) 6369 , c ) 5460 , d ) 7644 , e ) 6734 | d | divide(multiply(divide(multiply(7000, add(const_100, 4)), const_100), add(const_100, 5)), const_100) | find the amount on rs . 7000 in 2 years , the rate of interest being 4 % per first year and 5 % for the second year ? | "7000 * 104 / 100 * 105 / 100 = > 7644 answer : d" | a = 100 + 4
b = 7000 * a
c = b / 100
d = 100 + 5
e = c * d
f = e / 100
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a ) 300 , b ) 400 , c ) 600 , d ) 500 , e ) none of these | c | multiply(multiply(add(const_2, const_3), const_2), multiply(subtract(add(divide(14, const_100), const_1), add(const_1, divide(8, const_100))), 1000)) | a merchant has 1000 kg of sugar part of which he sells at 8 % profit and the rest at 18 % profit . he gains 14 % on the whole . the quantity sold at 18 % profit is | explanation : by the rule of alligation , we have profit % by selling 1 st part profit % by selling 2 nd part 8 18 net % profit 14 18 - 14 = 4 14 - 8 = 6 = > quantity of part 1 : quantity of part 2 = 4 : 6 = 2 : 3 total quantity is given as 1000 kg so quantity of part 2 ( quantity sold at 18 % profit ) = 1000 Γ 35 = 600 kg answer : option c | a = 2 + 3
b = a * 2
c = 14 / 100
d = c + 1
e = 8 / 100
f = 1 + e
g = d - f
h = g * 1000
i = b * h
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a ) rs . 950 , b ) rs . 1500 , c ) rs . 1000 , d ) rs . 1200 , e ) none of these | c | multiply(25, 40) | a trader sells 40 metres of cloth for rs . 8200 at a profit of rs . 25 per metre of cloth . how much profit will the trder earn on 40 metres of cloth ? | "explanation : sp of 1 metre cloth = 8200 / 40 = rs . 205 . cp of 1 metre cloth = rs . 205 β 25 = rs . 180 cp on 40 metres = 180 x 40 = rs . 7200 profit earned on 40 metres cloth = rs . 8200 β rs . 7200 = rs . 1000 . answer : option c" | a = 25 * 40
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a ) 23 , b ) 28 , c ) 27 , d ) 19 , e ) 24 | d | add(subtract(64, multiply(16, 3)), 3) | a batsman makes a score of 64 runs in the 16 th innings and thus increased his average by 3 . find his average after the 16 th inning ? | "let the average after the 16 th inning be p . so , the average after the 15 th inning will be ( p - 3 ) hence , 15 ( p - 30 ) + 64 = 16 p = > p = 19 . answer : d" | a = 16 * 3
b = 64 - a
c = b + 3
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a ) 40 , b ) 65 , c ) 73 , d ) 60 , e ) 80 | d | divide(2, divide(divide(const_1, multiply(2, add(const_2, 2))), const_2)) | when 2 is added to half of one - third of one - fifth of a number , the result is one - fifteenth of the number . find the number ? | "d 60 let the number be 2 + 1 / 2 [ 1 / 3 ( a / 5 ) ] = a / 15 = > 2 = a / 30 = > a = 60" | a = 2 + 2
b = 2 * a
c = 1 / b
d = c / 2
e = 2 / d
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a ) 627 , b ) 297 , c ) 127 , d ) 150 , e ) 181 | d | divide(multiply(const_100, divide(multiply(const_100, 225), add(25, const_100))), add(20, const_100)) | a sells a bicycle to b at a profit of 20 % . b sells it to c at a profit of 25 % . if c pays rs . 225 for it , the cost price of the bicycle for a is | explanation : 125 % of 120 % of a = 225 125 / 100 * 120 / 100 * a = 225 a = 225 * 2 / 3 = 150 . answer : d | a = 100 * 225
b = 25 + 100
c = a / b
d = 100 * c
e = 20 + 100
f = d / e
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a ) 16 , b ) 8 , c ) 4 , d ) 2.5 , e ) 1 | d | divide(divide(10, const_2), const_2) | if circles x and y have the same area and circle x has a circumference of 10 Ο , half of the radius of circle y is : | "x be radius of circle x y be radius of circle y given : pi * x ^ 2 = pi * y ^ 2 also , 2 * pi * x = 10 * pi x = 5 thus y = 5 y / 2 = 2.5 ans : d" | a = 10 / 2
b = a / 2
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a ) 5 , b ) 6 , c ) 8 , d ) 10 , e ) 50 | b | divide(subtract(multiply(multiply(2, multiply(2, multiply(2, multiply(2, 2)))), 2), const_4), const_10) | how many positive integers , from 2 to 100 , inclusive , are not divisible by even integers greater than 1 ? | "the no will be of the form 2 ^ n to achieve this . since any other form will have odd no in the prime factorization . hence we need to find solution of n for equation - 1 < 2 ^ n < 101 . 2 ^ 6 = 64 , 2 ^ 7 - 128 hence n can take values from 1 to 6 . hence ans - ( b ) 6" | a = 2 * 2
b = 2 * a
c = 2 * b
d = 2 * c
e = d * 2
f = e - 4
g = f / 10
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a ) 40 , b ) 48 , c ) 49 , d ) 55 , e ) 64 | c | add(divide(200, add(const_4, const_1)), divide(200, multiply(add(const_4, const_1), add(const_4, const_1)))) | how many terminating zeroes does 200 ! have ? | we must realize that the trailing zeros depend on the number of 10 s ( i . e . 10 s or its multiples ) in the factorial . 10 = 5 * 2 , so we are looking for 5 s and 2 s . by virtue of being the more frequent number , we do n ' t need to count the number of 2 s , because there will always be equal or more 2 s than 5 s in a factorial . so , essentially we are looking for 5 s and its exponents . start with finding 5 s . 200 / 5 = 40 . so we have 40 fives ( and 40 twos ) which means we ' ll have 40 zeros . if the answer was a decimal , such as 201 / 5 , round down . now look for two 5 s , that is 25 . so , 200 / 25 = 8 , so we have eight more zeros . lastly , 5 ^ 3 = 125 , and we have one such number in 200 ! , which gives us one more trailing zero . therefore , we have 40 + 8 + 1 = 49 trailing zeros . answer : c | a = 4 + 1
b = 200 / a
c = 4 + 1
d = 4 + 1
e = c * d
f = 200 / e
g = b + f
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a ) 65 , b ) 80 , c ) 90 , d ) 115 , e ) 120 | b | divide(multiply(40, 200), subtract(300, 200)) | two airplanes fly the same exact route from atlanta to chicago . plane a flies 200 mph while plane b flies 300 mph ( assume that simple model versus actual acceleration / deceleration near the airports ) . if plane a took off exactly 40 minutes before plane b , after how many minutes will plane b overtake plane a ? | we can use some form of the equation d = rt [ distance = rate * time ] rate of plane a : 200 mph rate of plane b : 300 mph plane a will be in the sky 40 minutes longer than plane b ( 2 / 3 of an hour ) time of plane a when b overtakes it : t + ( 2 / 3 ) use 2 / 3 of an hour since the rate is in hours time of plane b when b overtakes a : t at the time that b overtakes a , they will have travelled the same distance so rt is equal for each plane : 200 * ( t + 2 / 3 ) = 300 * t 200 t + 400 / 3 = 300 t 400 / 3 = 100 t 4 / 3 = t this is 1 and a third hour , so 60 minutes + 20 minutes = 80 minutes b | a = 40 * 200
b = 300 - 200
c = a / b
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a ) 2,858 , b ) 8,667 , c ) 21,429 , d ) 35,000 , e ) 70,000 | e | divide(multiply(divide(divide(multiply(add(add(const_3, const_10), 15), power(const_100, const_2)), subtract(15, 11)), multiply(const_4, const_2)), const_3), const_1000) | a certain manufacturer produces items for which the production costs consist of annual fixed costs totaling $ 130,000 and variables costs averaging $ 11 per item . if the manufacturer β s selling price per item is $ 15 , how many items the manufacturer produce and sell to earn an annual profit of $ 150,000 ? | "let the items manufactured or sold bex 130000 + 11 x = 15 x - 150000 4 x = 280000 x = 70000 ans : e" | a = 3 + 10
b = a + 15
c = 100 ** 2
d = b * c
e = 15 - 11
f = d / e
g = 4 * 2
h = f / g
i = h * 3
j = i / 1000
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a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 15 | e | sqrt(divide(450, const_2)) | the area of a parallelogram is 450 sq m and its altitude is twice the corresponding base . then the length of the base is ? | "2 x * x = 450 = > x = 15 answer : e" | a = 450 / 2
b = math.sqrt(a)
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a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 15 | e | subtract(subtract(divide(divide(multiply(32, 8), const_10), const_2), const_0_25), const_0_25) | a math teacher has 32 cards , each of which is in the shape of a geometric figure . half of the cards are rectangles , and a third of the cards are rhombuses . if 8 cards are squares , what is the maximum possible number of cards that re circles . | "a square is a special kind of rhombus ( sides are perpendicular ) a square is a special kind of rectangles ( sides with same length ) among the 32 cards with have : 15 rectangles 10 rhombus 8 squares among the 15 rectangles , there could be 8 special ones ( with sides of same length ) that are squares . that lets at least 7 rectangles that are not square . among the 10 rectangles , there could be 8 special ones ( with sides perpendicular ) that are squares . that lets at least 2 rhombus that are not square . we have 8 squares . so the minimum different cards that represent a square , a rhombus or a rectangle is 2 + 7 + 8 = 17 which means that the maximum number of circles that you could have is 32 - 17 = 15 answer ( e )" | a = 32 * 8
b = a / 10
c = b / 2
d = c - const_0_25
e = d - const_0_25
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a ) 2 , b ) 4 , c ) 10 , d ) 12 , e ) 15 | c | divide(add(15, 5), const_2) | in one hour , a boat goes 15 km along the stream and 5 km against the stream . the speed of the boat in still water ( in km / hr ) is : | "sol . speed in still water = 1 / 2 ( 15 + 5 ) kmph = 10 kmph . answer c" | a = 15 + 5
b = a / 2
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a ) 1394 , b ) 1294 , c ) 1194 , d ) 1094 , e ) none of them | a | subtract(multiply(multiply(const_100, divide(14, const_2)), const_2), multiply(const_2, const_3)) | find the least number which when divided by 20 , 2535 and 40 leaves remainders 14 , 1929 and 34 respectively . | here , ( 20 - 14 ) = 6 , ( 25 β 19 ) = 6 , ( 35 - 29 ) = 6 and ( 40 - 34 ) = 6 . therefore , required number = ( l . c . m . of 2025 , 3540 ) β 6 = 1394 . answer is a . | a = 14 / 2
b = 100 * a
c = b * 2
d = 2 * 3
e = c - d
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a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 7 | c | divide(28, add(multiply(3, 2), 4)) | bag a contains red , white and blue marbles such that the red to white marble ratio is 1 : 3 and the white to blue marble ratio is 2 : 3 . bag b contains red and white marbles in the ratio of 1 : 4 . together , the two bags contain 28 white marbles . how many red marbles could be in bag a ? | "# of red marbles in bag a can be either 2 or 6 . no 2 in the choices , so 6 . d . bag a : r : w : b = 2 : 6 : 9 bag b r : w = 1 : 4 6 x + 4 y = 28 i . e 3 x + 2 y = 14 x has to be odd to make an odd sum from the eq . x = 2 , y = 4 or x = 4 , y = 1 so r can be 2 x i . e 4 or 8 . answer c" | a = 3 * 2
b = a + 4
c = 28 / b
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a ) 17 , b ) 21 , c ) 22 , d ) 24 , e ) 25 | b | subtract(22, const_1) | when average age of 22 members are 0 , how many members greater than 0 ? | average of 22 numbers = 0 . sum of 22 numbers ( 0 x 22 ) = 0 . it is quite possible that 21 of these numbers may be positive and if their sum is a then 22 nd number is ( - a ) answer is 21 ( b ) | a = 22 - 1
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a ) 10 , b ) 9 , c ) 5 , d ) 15 , e ) 20 | a | multiply(1, 10) | if 10 spiders make 5 webs in 5 days , then how many days are needed for 1 spider to make 1 web ? | "explanation : let , 1 spider make 1 web in x days . more spiders , less days ( indirect proportion ) more webs , more days ( direct proportion ) hence we can write as ( spiders ) 10 : 1 ( webs ) 1 : 5 } : : x : 5 Γ’ β‘ β 10 Γ£ β 1 Γ£ β 5 = 1 Γ£ β 5 Γ£ β x Γ’ β‘ β x = 10 answer : option a" | a = 1 * 10
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a ) 100 , b ) 300 , c ) 400 , d ) 3,000 , e ) 4,000 | c | multiply(divide(12, subtract(99, 96)), const_100) | in a certain egg - processing plant , every egg must be inspected , and is either accepted for processing or rejected . for every 96 eggs accepted for processing , 4 eggs are rejected . if , on a particular day , 12 additional eggs were accepted , but the overall number of eggs inspected remained the same , the ratio of those accepted to those rejected would be 99 to 1 . how many e eggs does the plant process per day ? | "straight pluggin in for me . as usual , i started with c and got the answer . lets ' back calculate and see what we get let us consider eggs processed each day to be 400 so initial ratio of eggs processed and rejected is 96 : 4 or 24 : 1 so out of 400 eggs , there will be 384 eggs processed and 16 rejected . now if the no . of eggs inspected remain and 12 more eggs get accepted that means there e = 384 + 12 = 396 eggs accepted and 4 rejected . . . and the ratio will be 99 : 1 bingo . . . this is what the questions says . . . . its always a good idea to start with c ." | a = 99 - 96
b = 12 / a
c = b * 100
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a ) 12.18 , b ) 10.18 , c ) 10.12 , d ) 8.1 , e ) 8.2 | d | divide(2000, add(200, divide(200, const_4))) | ram and shakil run a race of 2000 m . first ram gives shakil a start of 200 m and beats him by 1 minute . next ram gives shakil a start of 6 min and is beaten by 1000 meters . find the time in minutes in which ram and shakil can run the race seperately . | let r be velocity of ram and s be the velocity of shakil then ( 2000 / r ) = ( 1800 / s ) - 60 . . . . . . ( 1 ) second case distance shakil travel in 6 min = 360 s again equating time for both 1000 / r = ( 2000 - 360 s ) / s . . . . . . . ( 2 ) solving ( 1 ) & ( 2 ) we get s = 20 / 6 putting value in first we get time = 480 sec = 8 min answer : d | a = 200 / 4
b = 200 + a
c = 2000 / b
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a ) 2 : 28 , b ) 5 : 28 , c ) 7 : 18 , d ) 1 : 18 , e ) 3 : 18 | d | divide(4000, 72000) | ravi and kavi start a business by investing Γ’ β ΒΉ 4000 and Γ’ β ΒΉ 72000 , respectively . find the ratio of their profits at the end of year . | "ratio of profit = ratio of investments = 4000 : 72000 = 1 : 18 answer : d" | a = 4000 / 72000
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a ) 9 % less , b ) 0.7 % less , c ) equal to each other , d ) 1 % more , e ) 9 % more | b | divide(const_100, subtract(multiply(const_100, const_100), multiply(add(const_100, 12), subtract(const_100, 12)))) | 108 . triangle a β s base is 12 % greater than the base of triangle b , and a β s height is 12 % less than the height of triangle b . the area of triangle a is what percent less or more than the area of triangle b ? | "wish the question specified that we are talking about corresponding height . base of a = 13 / 12 * base of b height of a = 11 / 12 * height of b area of a = ( 1 / 2 ) * base of a * height of a = 13 / 12 * 11 / 12 * area of b = 143 / 144 * area of b area of a is 0.7 % less than the area of b . answer ( b )" | a = 100 * 100
b = 100 + 12
c = 100 - 12
d = b * c
e = a - d
f = 100 / e
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a ) 166 , b ) 408 , c ) 400 , d ) 483 , e ) 566 | b | sqrt(subtract(power(multiply(283, const_2), const_2), power(multiply(204, 2), const_2))) | two airplanes take off from one airfield at noon . one flies due east at 204 miles per hour while the other flies directly northeast at 283 miles per hour . approximately how many miles apart are the airplanes at 2 p . m . ? | b in two hours : the plane flying east will be 408 miles away from airport . the other plane will be 566 miles away from airport . 566 / 408 = ~ 1.4 = ~ sqrt ( 2 ) this means that planes formed a right isocheles triangle = > sides of such triangles relate as 1 : 1 : sqrt ( 2 ) = > the planes are 408 miles apart . b | a = 283 * 2
b = a ** 2
c = 204 * 2
d = c ** 2
e = b - d
f = math.sqrt(e)
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a ) 1 : 0 , b ) 8 : 125 , c ) 1 : 7 , d ) 1 : 2 , e ) 1 : 1 | b | divide(power(2, const_3), power(const_3.0, 5)) | the triplicate ratio of 2 : 5 is ? | "23 : 53 = 8 : 125 answer : b" | a = 2 ** 3
b = 3 ** 0
c = a / b
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a ) 112 , b ) 56 , c ) 64 , d ) 49 , e ) 84 | b | multiply(multiply(multiply(const_10, const_10), subtract(const_10, const_1)), 4) | how many two - digit numbers that do not contain the digits 1 or 4 are there ? | "we can have 7 digits ( 2 , 3,5 , 6,7 , 8,9 ) for the first place ( tenth ' s place ) . and similarly 8 digits for the unit digit . ( 0,2 , 3,5 , 6,7 , 8,9 ) so in total 7 * 8 = 56 hence b" | a = 10 * 10
b = 10 - 1
c = a * b
d = c * 4
|
a ) 5 , b ) 48 , c ) 12 , d ) 15 , e ) 20 | b | divide(4, subtract(divide(10, multiply(add(5, const_1), 10)), divide(5, multiply(add(5, const_1), 10)))) | marla starts running around a circular track at the same time nick starts walking around the same circular track . marla completes 10 laps around the track per hour and nick completes 5 laps around the track per hour . how many minutes after marla and nick begin moving will marla have completed 4 more laps around the track than nick ? | maria ' s rate - 10 laps per hour - - > 10 / 60 laps / min nick ' s rate - 5 laps per hour - - > 5 / 60 laps / min lets set equations : 10 / 60 * t = 4 ( since maria had to run 4 laps before nick would start ) 5 / 60 * t = 0 ( hick has just started and has n ' t run any lap yet ) ( 10 / 60 - 5 / 60 ) * t = 4 - 0 ( since nick was chasing maria ) t = 48 min needed maria to run 4 laps answer : b | a = 5 + 1
b = a * 10
c = 10 / b
d = 5 + 1
e = d * 10
f = 5 / e
g = c - f
h = 4 / g
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a ) 70 , b ) 72 , c ) 75 , d ) 78 , e ) 80 | b | divide(add(add(multiply(75, 6), multiply(60, 4)), 30), add(6, 4)) | the average expenditure of a labourer for 6 months was 75 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income i | "income of 6 months = ( 6 Γ 75 ) β debt = 450 β debt income of the man for next 4 months = 4 Γ 60 + debt + 30 = 270 + debt β΄ income of 10 months = 720 average monthly income = 720 Γ· 10 = 72 answer b" | a = 75 * 6
b = 60 * 4
c = a + b
d = c + 30
e = 6 + 4
f = d / e
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a ) 1 % , b ) 4 % , c ) 6 % , d ) 8 % , e ) 9 % | a | multiply(subtract(const_1, subtract(const_2, add(divide(9, const_100), const_1))), const_100) | dick and jane each saved $ 2,500 in 1989 . in 1990 dick saved 9 percent more than in 1989 , and together he and jane saved a total of $ 5200 . approximately what percent less did jane save in 1990 than in 1989 ? | "1990 dick saved = $ 2725 jane saved = $ 2475 ( jane saved $ 25 less than she did the prior year ) jane saved approximately $ 25 / $ 2500 ( 1 % ) less in 1990 answer : a" | a = 9 / 100
b = a + 1
c = 2 - b
d = 1 - c
e = d * 100
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a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 60 % | c | divide(subtract(31, add(25, const_1)), subtract(divide(40, const_100), divide(add(25, const_1), const_100))) | seed mixture x is 40 % ryegrass and 60 % bluegrass by weight ; seed mixture y is 25 % ryegrass and 75 % fescue . if a mixture of x and y contains 31 % ryegrass , what percent of the weight of the mixture is from mixture x ? | "31 % is 6 % - points above 25 % and 9 % - points below 40 % . thus the ratio of mixture y to mixture x is 3 : 2 . the percent of mixture x is 2 / 5 = 40 % . the answer is c ." | a = 25 + 1
b = 31 - a
c = 40 / 100
d = 25 + 1
e = d / 100
f = c - e
g = b / f
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a ) 3 / 7 , b ) 11 / 35 , c ) 16 / 35 , d ) 31 / 70 , e ) 37 / 70 | b | add(multiply(divide(3, 7), subtract(2, divide(2, 3))), multiply(divide(const_2, 7), subtract(2, divide(3, 10)))) | in a tree , 3 / 7 of the birds are parrots while the rest are toucans . if 2 / 3 of the parrots are female and 7 / 10 of the toucans are female , what fraction of the birds in the tree are male ? | "let x be the number of birds in the tree . the fraction of birds that are male parrots is ( 1 / 3 ) ( 3 / 7 ) = 1 / 7 . the fraction of birds that are male toucans is ( 3 / 10 ) ( 4 / 7 ) = 6 / 35 . the total fraction of male birds is 1 / 7 + 6 / 35 = 11 / 35 . the answer is b ." | a = 3 / 7
b = 2 / 3
c = 2 - b
d = a * c
e = 2 / 7
f = 3 / 10
g = 2 - f
h = e * g
i = d + h
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a ) 2003 , b ) 2002 , c ) 2000 , d ) 2001 , e ) 2004 | c | subtract(10000, multiply(divide(8, 10), 10000)) | income and expenditure of a person are in the ratio 10 : 8 . if the income of the person is rs . 10000 , then find his savings ? | "let the income and the expenditure of the person be rs . 10 x and rs . 8 x respectively . income , 10 x = 10000 = > x = 1000 savings = income - expenditure = 10 x - 8 x = 2 x = 2 ( 1000 ) so , savings = rs . 2000 . answer : c" | a = 8 / 10
b = a * 10000
c = 10000 - b
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a ) 150 , b ) 200 , c ) 250 , d ) 350 , e ) none | a | subtract(multiply(12, 600), multiply(add(11, divide(9, 12)), 600)) | in a school with 600 students , the average age of the boys is 12 years and that of the girls is 11 years . if the average age of the school is 11 years 9 months , then the number of girls in the school is | "sol . let the number of grils be x . then , number of boys = ( 600 - x ) . then , ( 11 3 / 4 Γ 600 ) β 11 x + 12 ( 600 - x ) β x = 7200 - 7050 β 150 . answer a" | a = 12 * 600
b = 9 / 12
c = 11 + b
d = c * 600
e = a - d
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a ) 20 % , b ) 64.29 % , c ) 70 % , d ) 80 % , e ) 100 % | b | subtract(const_100, divide(subtract(const_100, 50), add(const_1, divide(40, const_100)))) | when sold at a 50 % discount , a sweater nets the merchant a 40 % profit on the wholesale cost at which he initially purchased the item . by what % is the sweater marked up from wholesale at its normal retail price ? | we should be careful about what are we measuring % on / what is the base . . let the marked up price = 100 . . selling price = 100 - 50 % of 100 = 50 . . profit = 40 % . . therefore the wholesale purchase cost = x . . . . 1.4 x = 50 or x = 35.71 . . . marked price was 100 so . . . so answer is 64.29 % . . b | a = 100 - 50
b = 40 / 100
c = 1 + b
d = a / c
e = 100 - d
|
a ) 34 % , b ) 28 % , c ) 22 % , d ) 18 % , e ) 8.5 % | b | multiply(divide(subtract(multiply(const_100, divide(16, const_100)), multiply(subtract(const_100, multiply(divide(const_1, const_4), const_100)), divide(12, const_100))), multiply(divide(const_1, const_4), const_100)), const_100) | one fourth of a solution that was 12 % sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight . the second solution was what percent sugar by weight ? | "instead of using complex calculations and remembering formulae , why dont u directly get to weighted average . 3 parts of 12 % + 1 part of x ( unknown ) % = 4 parts of 16 % = > x % = 64 % - 36 % = 28 % ans b it is ." | a = 16 / 100
b = 100 * a
c = 1 / 4
d = c * 100
e = 100 - d
f = 12 / 100
g = e * f
h = b - g
i = 1 / 4
j = i * 100
k = h / j
l = k * 100
|
a ) 40 , b ) 45 , c ) 50 , d ) 62 , e ) 2500 | d | divide(power(const_10, divide(4, const_2)), const_2) | a palindrome is a number that reads the same forward and backward , such as 144 . how many odd , 4 - digit numbers are palindromes ? | "a palindrome is a number that reads the same forward and backward . examples of four digit palindromes are 1221 , 4334 , 2222 etc you basically get to choose the first two digits and you repeat them in opposite order . say , you choose 45 as your first two digits . the next two digits are 54 and the number is 4554 . also , you need only odd palindromes . this means that you need an odd digit at the end i . e . 1 / 3 / 5 / 7 / 9 . this means that you need to start the number with an odd digit . only then will it end with an odd digit . in how many ways can you pick two digits such that the first one is an odd digit ? the first digit can be selected in 5 ways . ( 1 / 3 / 5 / 7 / 9 ) the second digit can be selected in 10 ways . ( 0 / 1 / 2 / 3 . . . 8 / 9 ) total = 5 * 12 = 62 ways d" | a = 4 / 2
b = 10 ** a
c = b / 2
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a ) 875 , b ) 405 , c ) 1425 , d ) 2025 , e ) 2500 | b | divide(multiply(subtract(30, const_3), 30), const_2) | a diagonal of a polygon is an segment between two non - adjacent vertices of the polygon . how many diagonals does a regular 30 - sided polygon have ? | "there ' s a direct formula for this . number of diagonals in a regular polygon = [ n * ( n - 3 ) ] / 2 , n = number of sides of the regular polygon . here , n = 30 . plugging it in , we get 405 diagonals ! answer ( b ) ." | a = 30 - 3
b = a * 30
c = b / 2
|
a ) 25 , b ) 31 , c ) 27 , d ) 29 , e ) 18 | e | add(subtract(66, multiply(17, 3)), 3) | a batsman makes a score of 66 runs in the 17 th inning and thus increases his averages by 3 . what is his average after 17 th inning ? | "let the average after 17 innings = x total runs scored in 17 innings = 17 x average after 16 innings = ( x - 3 ) total runs scored in 16 innings = 16 ( x - 3 ) total runs scored in 16 innings + 66 = total runs scored in 17 innings = > 16 ( x - 3 ) + 66 = 17 x = > 16 x - 48 + 66 = 17 x = > x = 18 answer is e ." | a = 17 * 3
b = 66 - a
c = b + 3
|
a ) 13.1 , b ) 43.1 , c ) 40.1 , d ) 23.1 , e ) 45.1 | b | divide(multiply(power(14, 2), power(5, 3)), 568) | find the value of e from ( 14 ) ^ 2 x 5 ^ 3 Γ£ Β· 568 = e . | given exp . = ( 14 ) ^ 2 x 5 ^ 3 Γ£ Β· 568 = e = 196 x 125 Γ£ Β· 568 43.1 b | a = 14 ** 2
b = 5 ** 3
c = a * b
d = c / 568
|
a ) 2878 , b ) 8000 , c ) 1000 , d ) 2997 , e ) 2701 | b | divide(power(power(5, const_2), const_3), power(5, const_3)) | a cube of side one meter length is cut into small cubes of side 5 cm each . how many such small cubes can be obtained ? | "along one edge , the number of small cubes that can be cut = 100 / 5 = 20 along each edge 5 cubes can be cut . ( along length , breadth and height ) . total number of small cubes that can be cut = 20 * 20 * 20 = 8000 answer : b" | a = 5 ** 2
b = a ** 3
c = 5 ** 3
d = b / c
|
a ) 2 : 3 , b ) 3 : 4 , c ) 1 : 30 , d ) 20 : 3 , e ) 30 : 7 | c | divide(multiply(0.025, const_100), multiply(0.75, const_100)) | if 0.75 of a number is equal to 0.025 of another number , the ratio of the numbers is : | "0.75 a = 0.025 b - > a / b = 0.025 / 0.75 = 25 / 750 = 1 / 30 : . a : b = 1 : 30 answer : c" | a = 0 * 25
b = 0 * 75
c = a / b
|
a ) 18 , b ) 16 , c ) 15 , d ) 17 , e ) 20 | e | divide(multiply(60, 30), add(60, 30)) | a company has two models of computers , model x and model y . operating at a constant rate , a model x computer can complete a certain task in 60 minutes and a model y computer can complete the same task in 30 minutes . if the company used the same number of each model of computer to complete the task in 1 minute , how many model x computers were used ? | "let ' s say 1 work is processing 60 gb of data . model x : 1 gb per min model y : 2 gb per min working together , 1 x and 1 y = 3 gb per min so , 20 times as many computers would work at 60 gb per min . so no . of x = 20 answer is e" | a = 60 * 30
b = 60 + 30
c = a / b
|
a ) 15 men , b ) 25 men , c ) 18 men , d ) 12 men , e ) 14 men | b | divide(multiply(add(40, 10), 5), 10) | a certain number of men can do a work in 40 days . if there were 5 men less it could be finished in 10 days more . how many men are there ? | "explanation : x * 40 = ( x - 5 ) * 50 5 x - 4 x = 25 x = 25 men answer : option b" | a = 40 + 10
b = a * 5
c = b / 10
|
a ) 50 % , b ) 32.5 % , c ) 37 % , d ) 37.5 % , e ) 40 % | a | multiply(divide(6, 12), const_100) | mike earns $ 12 per hour and phil earns $ 6 per hour . approximately how much less , as a percentage , does phil earn than mike per hour ? | "what % less of 12 is 6 let it be x % less , then = 12 ( 1 - x / 100 ) = 6 1 - x / 100 = 6 / 12 x = 100 / 2 x = 50 % ans a" | a = 6 / 12
b = a * 100
|
a ) 30 days , b ) 45 days , c ) 80 days , d ) 72 1 / 2 days , e ) 72 days | e | add(divide(const_1, 45), divide(const_1, 30)) | a can do a job in 45 days and b can do it in 30 days . a and b working together will finish four times the amount of work in - - - - - - - days ? | "1 / 45 + 1 / 30 = 5 / 90 = 1 / 18 18 / 1 = 18 * 4 = 72 days answer : e" | a = 1 / 45
b = 1 / 30
c = a + b
|
a ) 57 , b ) 51 , c ) 59 , d ) 55 , e ) 52 | a | add(40, divide(subtract(982, multiply(14, 40)), divide(multiply(14, add(const_100, 75)), const_100))) | a certain bus driver is paid a regular rate of $ 14 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 982 in total compensation , how many total hours did he work that week ? | "for 40 hrs = 40 * 14 = 560 excess = 982 - 560 = 422 for extra hours = . 75 ( 14 ) = 10.5 + 14 = 24.5 number of extra hrs = 422 / 24.5 = 17.2 = 17 approx . total hrs = 40 + 17 = 57 answer a" | a = 14 * 40
b = 982 - a
c = 100 + 75
d = 14 * c
e = d / 100
f = b / e
g = 40 + f
|
a ) 167 , b ) 243 , c ) 252 , d ) 336 , e ) none of these | b | multiply(lcm(lcm(lcm(3, 4), 5), 8), const_2) | the least number which when divided by 3 , 4 , 5 and 8 leaves a remainder 3 , but when divided by 9 leaves no remainder , is | "explanation : l . c . m of 3 , 4 , 5 , 8 = 120 therefore required number is of the form 120 k + 3 . least value of k for which ( 120 k + 3 ) is divisible by 9 is k = 2 therefore required number = ( 120 x 2 + 3 ) = 243 answer : b" | a = math.lcm(3, 4)
b = math.lcm(a, 5)
c = math.lcm(b, 8)
d = c * 2
|
a ) 23 , b ) 71 , c ) 27 , d ) 61 , e ) 81 | b | add(70, divide(multiply(5, 12), divide(180, 3))) | 70 + 5 * 12 / ( 180 / 3 ) = ? | "explanation : 70 + 5 * 12 / ( 180 / 3 ) = 70 + 5 * 12 / ( 60 ) = 70 + ( 5 * 12 ) / 60 = 70 + 1 = 71 . answer : b" | a = 5 * 12
b = 180 / 3
c = a / b
d = 70 + c
|
a ) 54 , b ) 32 , c ) 75 , d ) 28 , e ) 16 | e | sqrt(divide(multiply(square_area(4), 8), inverse(const_2))) | the length of the rectangular field is double its width . inside the field there is square shaped pond 4 m long . if the area of the pond is 1 / 8 of the area of the field . what is the length of the field ? | "a / 8 = 4 * 4 = > a = 4 * 4 * 8 x * 2 x = 4 * 4 * 8 x = 8 = > 2 x = 16 answer : e" | a = square_area * (
b = a / 8
c = 1/(2)
d = math.sqrt(b)
|
a ) 27.5 days , b ) 15.5 days , c ) 17.5 days , d ) 16.5 days , e ) 18.5 days | c | divide(multiply(25, 14), 20) | 14 men can complete a piece of work in 25 days . in how many days can 20 men complete that piece of work ? | "14 * 25 = 20 * x = > x = 17.5 days answer : c" | a = 25 * 14
b = a / 20
|
a ) 276 , b ) 271 , c ) 200 , d ) 278 , e ) 268 | c | multiply(divide(multiply(40, const_1), subtract(50, 40)), 50) | a train leaves mumabai at 9 am at a speed of 40 kmph . after one hour , another train leaves mumbai in the same direction as that of the first train at a speed of 50 kmph . when and at what distance from mumbai do the two trains meet ? | "when the second train leaves mumbai the first train covers 40 * 1 = 40 km so , the distance between first train and second train is 40 km at 10.00 am time taken by the trains to meet = distance / relative speed = 40 / ( 50 - 40 ) = 4 hours so , the two trains meet at 2 p . m . the two trains meet 4 * 50 = 200 km away from mumbai . answer : c" | a = 40 * 1
b = 50 - 40
c = a / b
d = c * 50
|
a ) 30 / 25 , b ) 28 / 13 , c ) 28 / 29 , d ) 8 / 13 , e ) none of these | d | divide(368, 598) | reduce 368 / 598 to the lowest terms | explanation : we can do it easily by in two steps step 1 : we get the hcf of 368 and 598 which is 46 step 2 : divide both by 46 , we will get the answer 8 / 13 answer : option d | a = 368 / 598
|
a ) 70000 , b ) 60000 , c ) 88000 , d ) 90000 , e ) 50000 | c | add(add(multiply(divide(5000, 9000), 36000), multiply(divide(8000, 9000), 36000)), 36000) | a , b and c started a partnership business by investing rs . 5000 , rs . 8000 , rs . 9000 respectively . at the end of the year , the profit were distributed among them . if c ' s share of profit is 36000 , what is the total profit ? | "a : b : c = 5000 : 8000 : 9000 = 5 : 8 : 9 let total profit = p then p Γ£ β 9 / 22 = 36000 p = ( 36000 Γ£ β 22 ) / 9 = 88000 answer is c ." | a = 5000 / 9000
b = a * 36000
c = 8000 / 9000
d = c * 36000
e = b + d
f = e + 36000
|
a ) 1 and 8 , b ) 2 and 6 , c ) 0 and 9 , d ) 2 and 7 , e ) 2 and 6 | e | add(multiply(2, const_10), divide(add(37, 5), 7)) | 5 n + 2 > 12 and 7 n - 5 < 37 ; n must be between which numbers ? | "5 n > 10 - - > n > 2 7 n < 42 - - > n < 6 2 < n < 6 answer : e" | a = 2 * 10
b = 37 + 5
c = b / 7
d = a + c
|
a ) rs . 500 , b ) rs . 600 , c ) rs . 650 , d ) rs . 720 , e ) none | b | multiply(multiply(100, divide(add(add(multiply(const_10, const_1000), multiply(const_4, const_1000)), multiply(const_4, 100)), add(100, multiply(100, divide(20, 100))))), divide(5, 100)) | a man invested rs . 14,400 in rs . 100 shares of a company at 20 % premium . if the company declares 5 % dividend at the end of the year , then how much does he get ? | "solution number of shares = ( 14400 / 120 ) = 120 . face value = rs . ( 100 x 120 ) = rs . 12000 . annual income = rs . ( 5 / 100 x 12000 ) = rs . 600 . answer b" | a = 10 * 1000
b = 4 * 1000
c = a + b
d = 4 * 100
e = c + d
f = 20 / 100
g = 100 * f
h = 100 + g
i = e / h
j = 100 * i
k = 5 / 100
l = j * k
|
a ) 4 / 5 , b ) 5 / 4 , c ) 3 / 2 , d ) 4 / 7 , e ) 2 / 3 | d | divide(sqrt(16), sqrt(49)) | two isosceles triangles have equal vertical angles and their areas are in the ratio 16 : 49 . find the ratio of their corresponding heights . | "we are basically given that the triangles are similar . in two similar triangles , the ratio of their areas is the square of the ratio of their sides and also , the square of the ratio of their corresponding heights . therefore , area / area = height ^ 2 / height ^ 2 = 16 / 49 - - > height / height = 4 / 7 . answer : d ." | a = math.sqrt(16)
b = math.sqrt(49)
c = a / b
|
a ) 65 % , b ) 83 % , c ) 27 % , d ) 42 % , e ) 75 % | e | subtract(const_100, multiply(divide(add(50, const_100), add(100, const_100)), const_100)) | two numbers are respectively 50 % and 100 % more than a third number . the percentage that is first of the second is ? | "i ii iii 150 200 100 200 - - - - - - - - - - 150 100 - - - - - - - - - - - ? = > 75 % answer : e" | a = 50 + 100
b = 100 + 100
c = a / b
d = c * 100
e = 100 - d
|
a ) 39 , b ) 38 , c ) 40 , d ) 41 , e ) 34 | d | divide(divide(300, 5280), multiply(5, divide(1, const_3600))) | if an object travels 300 feet in 5 seconds , what is the object ' s approximate speed in miles per hour ? ( note : 1 mile = 5280 feet ) | "1 mile = 5280 feet = > 1 feet = 1 / 5280 miles if the object travels 300 feet in 5 sec then it travels 300 / 5 * 60 * 60 feet in 1 hour ( 1 hr = 60 min * 60 sec ) = 3600 * 60 feet in 1 hour = 216000 feet in 1 hr = 216000 / 5280 miles in 1 hour = 21600 / 528 miles / hr ~ 41 miles / hr answer - d" | a = 300 / 5280
b = 1 / 3600
c = 5 * b
d = a / c
|
a ) rs . 1020.80 , b ) rs . 1025 , c ) rs . 1061 , d ) data inadequate , e ) none of these | c | add(875, divide(multiply(multiply(875, add(divide(multiply(subtract(956, 875), const_100), multiply(875, 3)), 4)), 3), const_100)) | rs . 875 becomes rs . 956 in 3 years at a certain rate of simple interest . if the rate of interest is increased by 4 % , what amount will rs . 875 become in 3 years ? | "solution s . i . = rs . ( 956 - 875 ) = rs . 81 rate = ( 100 x 81 / 875 x 3 ) = 108 / 35 % new rate = ( 108 / 35 + 4 ) % = 248 / 35 % new s . i . = rs . ( 875 x 248 / 35 x 3 / 100 ) rs . 186 . β΄ new amount = rs . ( 875 + 186 ) = rs . 1061 . answer c" | a = 956 - 875
b = a * 100
c = 875 * 3
d = b / c
e = d + 4
f = 875 * e
g = f * 3
h = g / 100
i = 875 + h
|
a ) 9 , b ) 3 , c ) 0.3 , d ) 0.03 , e ) none of these | c | multiply(0.003, 0.0003) | 0.003 * ? * 0.0003 = 0.00000027 | "explanation : 0.003 * ? * 0.0003 = 0.00000027 3 / 1000 * ? * 3 / 10000 = 3 / 1000 * 3 / 1000 * 3 / 100 ? = 3 / 10 = 0.3 answer is c" | a = 0 * 3
|
a ) rs . 1200 , b ) rs . 1500 , c ) rs . 2040 , d ) rs . 2000 , e ) none of these | c | multiply(divide(divide(multiply(15, const_2), 10), add(divide(multiply(15, const_2), 10), divide(multiply(15, const_2), 15))), 3400) | a alone can finish a work in 10 days which b alone can finish in 15 days . if they work together and finish it , then out of a total wages of rs . 3400 , a will get : | "explanation : ratio of working days of a : b = 10 : 15 therefore , their wages ratio = reverse ratio = 15 : 10 therefore , a will get 15 units of ratio total ratio = 25 1 unit of ratio = 3000 / 25 = 136 so , a β s amount = 120 Γ 15 = rs . 2040 . answer : option c" | a = 15 * 2
b = a / 10
c = 15 * 2
d = c / 10
e = 15 * 2
f = e / 15
g = d + f
h = b / g
i = h * 3400
|
a ) 18 , b ) 40 , c ) 21 , d ) 43 , e ) 28 | a | subtract(add(power(19, 19), 19), multiply(20, floor(divide(add(power(19, 19), 19), 20)))) | what will be the reminder when ( 19 ^ 19 + 19 ) is divided by 20 ? | "( x ^ n + 1 ) will be divisible by ( x + 1 ) only when n is odd ; ( 19 ^ 19 + 1 ) will be divisible by ( 19 + 1 ) ; ( 19 ^ 19 + 1 ) + 18 when divided by 20 will give 18 as remainder . correct option : a" | a = 19 ** 19
b = a + 19
c = 19 ** 19
d = c + 19
e = d / 20
f = math.floor(e)
g = 20 * f
h = b - g
|
a ) 0.4 % , b ) 0.8 % , c ) 2 % , d ) 4 % , e ) 8 % | b | subtract(subtract(12, 10), divide(12, 10)) | a couple who own an appliance store discover that if they advertise a sales discount of 10 % on every item in the store , at the end of one month the number of total items sold increases 12 % . their gross income from sales for one month increases by what percent ? | "let p be the original price and let x be the number of items sold originally . the original income is p * x . after the changes , the income is 0.9 p * 1.12 x = 1.008 * ( p * x ) , an increase of 0.8 % . the answer is b ." | a = 12 - 10
b = 12 / 10
c = a - b
|
a ) 8 % , b ) 15 % , c ) 40 % , d ) 52 % , e ) 56 % | c | multiply(divide(3, 20), const_100) | a pharmaceutical company received $ 3 million in royalties on the first $ 20 million in sales of and then $ 9 million in royalties on the next $ 102 million in sales . by approximately what percentage did the ratio of royalties to sales decrease from the first $ 20 million in sales to the next $ 102 million in sales ? | "( 9 / 102 ) / ( 3 / 20 ) = 30 / 54 = 58,5 % it means that 9 / 108 represents only 58,5 % . therefore a decrease of 45 % . answer c" | a = 3 / 20
b = a * 100
|
a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25 | c | subtract(subtract(55, divide(60, const_3)), divide(60, const_3)) | the perimeter of an equilateral triangle is 60 . if one of the sides of the equilateral triangle is the side of an isosceles triangle of perimeter 55 , then how long is the base of isosceles triangle ? | "the base of the isosceles triangle is 55 - 20 - 20 = 15 units the answer is c ." | a = 60 / 3
b = 55 - a
c = 60 / 3
d = b - c
|
a ) a ) 140 , b ) b ) 150 , c ) c ) 120 , d ) d ) 200 , e ) e ) 220 | c | divide(multiply(divide(10, multiply(multiply(divide(const_1, const_4), divide(const_1, const_3)), divide(const_2, add(const_2, const_3)))), 40), const_100) | one fourth of one third of two fifth of a number is 10 . what will be 40 % of that number | "explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 10 then x = 10 * 30 = 300 40 % of 300 = 120 answer : option c" | a = 1 / 4
b = 1 / 3
c = a * b
d = 2 + 3
e = 2 / d
f = c * e
g = 10 / f
h = g * 40
i = h / 100
|
a ) 0 , b ) 5 , c ) 10 , d ) 15 , e ) 22 | e | subtract(multiply(reminder(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000), 52), const_1), 52), 11,0), const_10), divide(subtract(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000), 52), const_1), 52), 11,0), reminder(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000), 52), const_1), 52), 11,0), const_10)), const_10)), const_10) | in the number 11,0 ab , a and b represent the tens and units digits , respectively . if 11,0 ab is divisible by 52 , what is the greatest possible value of b Γ a ? | "you should notice that 52 * 2 = 110 so 11,000 is divisible by 55 : 55 * 200 = 11,000 ( or you can notice that 11,000 is obviously divisible by both 5 and 11 so by 55 ) - - > b * a = 0 * 0 = 0 . next number divisible by 55 is 11,000 + 55 = 11,055 : b * a = 5 * 5 = 22 ( next number wo n ' t have 110 as the first 3 digits so we have only two options 0 and 25 ) . answer : e . ! please post ps questions in the ps subforum : gmat - problem - solving - ps - 140 / please post ds questions in the ds subforum : gmat - data - sufficiency - ds - 141 / no posting of ps / ds questions is allowed in the mainmath forum . e" | a = 11 ** 0
b = a + 1000
c = b / 52
d = c + 1
e = d * 52
f = reminder * (
g = f - 10
|
a ) 10 , b ) 15 , c ) 12.5 , d ) 22.5 , e ) 21 | d | divide(add(add(add(8, const_1), add(add(const_4.0, const_1), const_2)), add(subtract(5, 8), subtract(5, const_2))), 8) | find the average of first 8 multiples of 5 ? | "average = ( 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 ) / 8 = 22.5 answer is d" | a = 8 + 1
b = 4 + 0
c = b + 2
d = a + c
e = 5 - 8
f = 5 - 2
g = e + f
h = d + g
i = h / 8
|
a ) 3 , b ) 6 , c ) 9 , d ) 8 , e ) 7 | e | multiply(divide(subtract(subtract(45, const_1), multiply(subtract(16, const_1), const_2)), subtract(multiply(subtract(16, const_1), const_4), multiply(subtract(16, const_1), const_2))), subtract(16, const_1)) | a luxury liner , queen marry ii , is transporting several cats as well as the crew ( sailors , a cook , and one - legged captain ) to a nearby port . altogether , these passengers have 16 heads and 45 legs . how many cats does the ship host ? | "a ' s + co + ca + cats = 16 . sa ' s + 1 + 1 + cats = 16 or sa ' s + cats = 14 . sa ' s ( 2 ) + 2 + 1 + cats * 4 = 45 sa ' s * 2 + cats * 4 = 42 or sa ' s + cats * 2 = 21 or 14 - cats + cat * 2 = 21 then cats = 7 e" | a = 45 - 1
b = 16 - 1
c = b * 2
d = a - c
e = 16 - 1
f = e * 4
g = 16 - 1
h = g * 2
i = f - h
j = d / i
k = 16 - 1
l = j * k
|
a ) 913 , b ) 825 , c ) 330 , d ) 582 , e ) 465 | a | divide(multiply(2310, 83), 210) | the lcm of two numbers is 2310 and hcf is 83 . if one of the numbers is 210 . then what is the other number ? | "first number * second number = lcm * hcf other number = 2310 * 83 / 210 = 11 * 83 = 913 answer : a" | a = 2310 * 83
b = a / 210
|
a ) 55 , b ) 60 , c ) 70 , d ) 50 , e ) 65 | b | subtract(divide(multiply(2, multiply(multiply(100, 25), 8)), multiply(25, 10)), 100) | nhai employs 100 men to build a highway of 2 km in 50 days working 8 hours a day . if in 25 days , they completed 1 / 3 part of work . than how many more employees should nhai hire to finish it in time working 10 hours a day ? | explanation : the given problem can be written in a tabular form like below : - men days hour work planned 100 50 8 1 actual 100 25 8 1 / 3 remaining ? 25 10 2 / 3 we can apply chain rule now . total men required to complete the remaining work = 100 Γ ( 25 / 25 ) Γ ( 8 / 10 ) Γ ( 2 / 3 ) / ( 1 / 3 ) = 160 . so additional men required = 160 - 100 = 60 . answer is b | a = 100 * 25
b = a * 8
c = 2 * b
d = 25 * 10
e = c / d
f = e - 100
|
a ) 3 , b ) 1 , c ) 16 , d ) 11 , e ) 15 | b | subtract(9671, multiply(floor(divide(9671, 2)), 2)) | what least number must be subtracted from 9671 so that the remaining number is divisible by 2 ? | "on dividing 9671 by 2 , we get remainder = 1 . required number be subtracted = 1 answer : b" | a = 9671 / 2
b = math.floor(a)
c = b * 2
d = 9671 - c
|
a ) 18500 , b ) 18200 , c ) 17500 , d ) 16500 , e ) 13500 | b | multiply(add(divide(200, 90), 200), 90) | find the value of ( 200 + 200 / 90 ) Γ 90 | "( 18000 + 200 ) / 90 * 90 = 18200 answer : b" | a = 200 / 90
b = a + 200
c = b * 90
|
a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 52 | e | divide(divide(multiply(130, const_4), const_10), const_2) | mr . loyd wants to fence his square shaped land of 130 sqft each side . if a pole is laid every 10 ft how many poles do he need ? | "if each side is 120 feet . . then total perimeter is 130 * 4 = 520 poles every 10 feet hence no of poles = 520 / 10 = 52 answer : e" | a = 130 * 4
b = a / 10
c = b / 2
|
a ) 20,000 , b ) 176470 , c ) 10000 , d ) 13000 , e ) 56600 | b | add(subtract(80, const_10), add(add(divide(600000, const_100), add(add(multiply(const_1000, const_10), divide(600000, 4)), multiply(const_1000, const_10))), multiply(const_100, const_4))) | a beekeeper has 600000 bees living in 4 beehives . among these beehives , the sizes of the beehives are proportional to the numbers of bees living in them . no beehive has less than 80 % the number of bees of any other beehive . what is the maximum possible number of bees that the greatest beehive could have ? | x be the # of bees in the hive with the max # of bees . let other hives have same # of bees . so , x + 3 * ( 0.8 ) * x = 600,000 = > 3.4 x = 600,000 = > x = 176470 . b is the answer . | a = 80 - 10
b = 600000 / 100
c = 1000 * 10
d = 600000 / 4
e = c + d
f = 1000 * 10
g = e + f
h = b + g
i = 100 * 4
j = h + i
k = a + j
|
a ) 22 , b ) 38 , c ) 60 , d ) 88 , e ) 90 | e | divide(multiply(36, 25), 10) | if 36 men can do a piece of work in 25 hours , in how many hours will 10 men do it ? | explanation : let the required no of hours be x . then less men , more hours ( indirect proportion ) \ inline \ fn _ jvn \ therefore 10 : 36 : : 25 : x \ inline \ fn _ jvn \ leftrightarrow ( 10 x x ) = ( 36 x 25 ) \ inline \ fn _ jvn \ leftrightarrow \ inline \ fn _ jvn x = \ frac { 36 \ times 25 } { 10 } = 90 hence , 10 men can do it in 90 hours . answer : e ) 90 | a = 36 * 25
b = a / 10
|
a ) 2.8 % , b ) 3.6 % , c ) 4.4 % , d ) 5.2 % , e ) 6.0 % | d | multiply(divide(add(multiply(50, divide(4, const_100)), multiply(40, divide(8, const_100))), const_100), const_100) | of the total amount that jill spent on a shopping trip , excluding taxes , she spent 50 percent on clothing , 10 percent on food , and 40 percent on other items . if jill paid a 4 percent tax on the clothing , no tax on the food , and an 8 percent tax on all other items , then the total tax that she paid was what percent of the total amount that she spent , excluding taxes ? | "let amount spent by jill = 100 clothing = 50 , food = 10 , others = 40 tax on clothing = 2 tax on others = 3.2 percentage = 5.2 / 100 = 5.2 % answer : d" | a = 4 / 100
b = 50 * a
c = 8 / 100
d = 40 * c
e = b + d
f = e / 100
g = f * 100
|
a ) 18 , b ) 25 , c ) 27 , d ) 29 , e ) 31 | c | add(subtract(5, 4), const_1) | if the average ( arithmetic mean ) of 4 x - 5 , 8 x - 6 , and 9 x + 5 is 279 , what is the value of x ? | "am of 4 x - 5 , 8 x - 6 , and 9 x + 5 = 4 x - 5 + 8 x - 6 + 9 x + 5 / 3 = 31 x / 3 given that 31 x / 3 = 279 x = 837 / 31 = 27 x = 27 answer is c" | a = 5 - 4
b = a + 1
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | c | subtract(multiply(divide(34, divide(add(const_100, 70), const_100)), const_2), 34) | a retailer bought a hat at wholesale and marked it up 70 % to its initial price of $ 34 . by how many more dollars does he need to increase the price to achieve a 100 % markup ? | let x be the wholesale price . then 1.7 x = 34 and x = 34 / 1.7 = 20 . to achieve a 100 % markup , the price needs to be $ 40 . the retailer needs to increase the price by $ 6 more . the answer is c . | a = 100 + 70
b = a / 100
c = 34 / b
d = c * 2
e = d - 34
|
a ) 1 / 16 , b ) 37 / 256 , c ) 1 / 2 , d ) 219 / 256 , e ) 15 / 16 | e | add(divide(subtract(const_1, add(add(power(divide(const_1, const_2), 8), multiply(8, power(divide(const_1, const_2), 8))), multiply(multiply(8, const_3), power(divide(const_1, const_2), 8)))), const_10), subtract(const_1, add(add(power(divide(const_1, const_2), 8), multiply(8, power(divide(const_1, const_2), 8))), multiply(multiply(8, const_3), power(divide(const_1, const_2), 8))))) | an exam consists of 8 true / false questions . brian forgets to study , so he must guess blindly on each question . if any score above 90 % is a passing grade , what is the probability that brian passes ? | "if you have 8 t or f and brian is going to guess then each question he has a 50 % chance of getting correct . if a passing score is 70 % it means brian needs to get 6 / 8 = 75 % , 7 / 8 = 87.5 % , or 8 / 8 = 100 % to pass . each is a possibility . if brian gets a 5 / 8 ( = 62.5 % ) or below he fails . so first figure out the number of ways that brian can get 6 out of 8 , 7 out of 8 , and 8 out of 8 questions correct . which is 8 choose 6 , equals is 28 , 8 choose 7 , equals 8 , and 8 choose 8 , equals 1 . this sums to 37 . the number of possible questions outcomes - the sum of 8 choose 8 , 7 choose 8 , 6 choose 8 β¦ . 2 choose 8 , 1 choose 8 , and 0 choose 8 is 256 , so the chance of him passing is 15 / 16 . e" | a = 1 / 2
b = a ** 8
c = 1 / 2
d = c ** 8
e = 8 * d
f = b + e
g = 8 * 3
h = 1 / 2
i = h ** 8
j = g * i
k = f + j
l = 1 - k
m = l / 10
n = 1 / 2
o = n ** 8
p = 1 / 2
q = p ** 8
r = 8 * q
s = o + r
t = 8 * 3
u = 1 / 2
v = u ** 8
w = t * v
x = s + w
y = 1 - x
z = m + y
|
a ) 420 , b ) 600 , c ) 220 , d ) 400 , e ) none of these | b | add(420, multiply(420, divide(30, const_100))) | a fruit seller had some oranges . he sells 30 % oranges and still has 420 oranges . how many oranges he had originally ? | "explanation : he sells 30 % of oranges and still there are 420 oranges remaining = > 70 % of oranges = 420 β ( 70 Γ total oranges ) / 100 = 420 β total oranges / 100 = 6 β total oranges = 6 Γ 100 = 600 answer : option b" | a = 30 / 100
b = 420 * a
c = 420 + b
|
a ) 150 , b ) 997 , c ) 266 , d ) 300 , e ) 271 | d | divide(add(70, 20), divide(30, const_100)) | 70 is subtracted from 30 % of a number , the result is 20 . find the number ? | ( 30 / 100 ) * x β 70 = 20 3 x = 900 x = 300 answer : d | a = 70 + 20
b = 30 / 100
c = a / b
|
a ) 9320 , b ) 96288 , c ) 26667 , d ) 1662 , e ) 2882 | a | add(multiply(multiply(add(divide(0.5, const_100), divide(divide(subtract(9200, 8000), 3), 8000)), 8000), 3), 8000) | sonika deposited rs . 8000 which amounted to rs . 9200 after 3 years at simple interest . had the interest been 0.5 % more . she would get how much ? | "( 8000 * 3 * 0.5 ) / 100 = 120 9200 - - - - - - - - 9320 answer : a" | a = 0 / 5
b = 9200 - 8000
c = b / 3
d = c / 8000
e = a + d
f = e * 8000
g = f * 3
h = g + 8000
|
a ) a ) 258 , b ) b ) 65 , c ) c ) 58 , d ) d ) 62 , e ) e ) 48 | a | subtract(add(multiply(6, 88), multiply(6, 65)), multiply(11, 60)) | the average of 11 numbers is 60 . out of 11 numbers the average of first 6 no . is 88 , and last 6 numbers is 65 then find 6 th number ? | "6 th number = sum of 1 st 6 no . s + sum of last 6 no . s - sum of 11 no . s answer = 6 * 88 + 6 * 65 - 11 * 60 = 258 answer is a" | a = 6 * 88
b = 6 * 65
c = a + b
d = 11 * 60
e = c - d
|
a ) 225 sq m , b ) 169 sq m , c ) 128 sq m , d ) 267 sq m , e ) 231 sq m | b | divide(square_area(13), const_2) | what is the area of a square field whose sides have a length of 13 meters ? | "13 * 13 = 169 sq m the answer is b ." | a = square_area / (
|
a ) 96.9 , b ) 96.3 , c ) 94.2 , d ) 96.7 , e ) 96.21 | c | add(subtract(100, 6), divide(1, 5)) | what is the cp of rs 100 stock at 6 discount , with 1 / 5 % brokerage ? | "explanation : use the formula , cp = 100 Γ’ β¬ β discount + brokerage % cp = 100 - 6 + 1 / 5 94.2 thus the cp is rs 94.2 . answer : c" | a = 100 - 6
b = 1 / 5
c = a + b
|
['a ) 228', 'b ) 148', 'c ) 992', 'd ) 772', 'e ) 222'] | b | add(multiply(multiply(const_2, const_4), 5), add(multiply(multiply(const_2, 6), const_4), multiply(6, multiply(const_2, 5)))) | the edges of cuboid are 4 cm ; 5 cm and 6 cm . find its surface area ? | 2 ( 4 * 5 + 5 * 6 + 4 * 6 ) = 148 answer : b | a = 2 * 4
b = a * 5
c = 2 * 6
d = c * 4
e = 2 * 5
f = 6 * e
g = d + f
h = b + g
|
a ) 10 , b ) 12 , c ) 20 , d ) 30 , e ) 15 | a | inverse(add(divide(const_1, 15), divide(const_1, 30))) | x does a work in 15 days . y does the same work in 30 days . in how many days they together will do the same work ? | "x ' s 1 day ' s work = 1 / 15 y ' s 1 day ' s work = 1 / 30 ( x + y ) ' s 1 day ' s work = ( 1 / 15 + 1 / 30 ) = 1 / 10 both together will finish the work in 10 days . correct option is a" | a = 1 / 15
b = 1 / 30
c = a + b
d = 1/(c)
|
a ) 16 sec , b ) 13 sec , c ) 17 sec , d ) 21 sec , e ) 23 sec | b | divide(multiply(120, const_2), add(speed(120, 20), speed(120, 10))) | two trains of equal lengths take 10 sec and 20 sec respectively to cross a telegraph post . if the length of each train be 120 m , in what time will they cross other travelling in opposite direction ? | "speed of the first train = 120 / 10 = 12 m / sec . speed of the second train = 120 / 20 = 6 m / sec . relative speed = 12 + 6 = 18 m / sec . required time = ( 120 + 120 ) / 18 = 13 sec . answer : b" | a = 120 * 2
b = speed + (
c = a / b
|
a ) 2400 , b ) 2550 , c ) 5050 , d ) 6275 , e ) 11325 | a | multiply(subtract(104, 4), add(divide(subtract(50, 4), const_2), const_1)) | set a contains all the even numbers between 4 and 50 inclusive . set b contains all the even numbers between 104 and 150 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ? | "set a contains 4 , 6 . . . 50 set b contains 104 , 106 . . . 150 number of terms in each set = 24 difference between corresponding terms in set a and b = 100 difference between sum of set b and set a = 100 * 24 = 2400 answer a" | a = 104 - 4
b = 50 - 4
c = b / 2
d = c + 1
e = a * d
|
a ) 0.2 , b ) 1.2 , c ) 1.8 , d ) 2.2 , e ) 4.0 | d | subtract(8.2, subtract(floor(8.2), const_1)) | for any number y , y * is defined as the greatest positive even integer less than or equal to y . what is the value of 8.2 β 8.2 * ? | "since y * is defined as the greatest positive even integer less than or equal to y , then 8.2 * = 4 ( the greatest positive even integer less than or equal to 5.2 is 4 ) . hence , 8.2 β 8.2 * = 8.2 - 6 = 2.2 answer : d ." | a = math.floor(8, 2)
b = a - 1
c = 8 - 2
|
a ) 10 , b ) 100 , c ) 1000 , d ) 10000 , e ) none of these | a | multiply(1000, 10) | ( 1000 ) 7 Γ· ( 10 ) 20 = ? | "explanation : = ( 103 ) 7 / ( 10 ) 20 = ( 10 ) 21 / ( 10 ) 20 = 10 ( 1 ) = 10 option a" | a = 1000 * 10
|
a ) 74 , b ) v = 75 , c ) v = 175 , d ) v = 680 , e ) 690 | b | add(multiply(34, const_2), divide(subtract(multiply(34, const_2), multiply(8, 5)), subtract(5, const_1))) | a number when divided by 5 gives a number which is 8 more than the remainder obtained on dividing the same number by 34 . such a least possible number v is | i solved this question by plugging in numbers from the answer choices . a . ) 74 starting with answer choice a , i immediately eliminated it because 74 is not even divisible by 5 . b . ) 75 i divide 75 / 5 and get 15 as an answer . i divide 75 / 34 and get a remainder of 7 . 15 - 7 = 8 so i know the correct answer isb | a = 34 * 2
b = 34 * 2
c = 8 * 5
d = b - c
e = 5 - 1
f = d / e
g = a + f
|
a ) 55 , b ) 50 , c ) 40 . , d ) 30 . , e ) 25 . | b | add(multiply(divide(subtract(const_100, 10), const_100), 50), multiply(divide(10, const_100), 50)) | in the hillside summer camp there are 50 children . 80 % of the children are boys and the rest are girls . the camp administrator decided to make the number of girls only 10 % of the total number of children in the camp . how many more boys must she bring to make that happen ? | "given there are 50 students , 80 % of 50 = 40 boys and remaining 10 girls . now here 80 % are boys and 20 % are girls . now question is asking about how many boys do we need to add , to make the girls percentage to 20 or 20 % . . if we add 50 to existing 40 then the count will be 90 and the girls number will be 10 as it . now boys are 90 % and girls are 10 % . ( out of 100 students = 90 boys + 10 girls ) . imo option b is correct ." | a = 100 - 10
b = a / 100
c = b * 50
d = 10 / 100
e = d * 50
f = c + e
|
a ) rs . 1200 , b ) rs . 1300 , c ) rs . 1500 , d ) rs . 4000 , e ) none of these | d | multiply(multiply(8, const_1000), divide(subtract(6, 5), subtract(8, 6))) | a man invests rs . 8,000 at the rate of 5 % per annum . how much more should he invest at the rate of 8 % , so that he can earn a total of 6 % per annum ? | explanation : interest on rs . 8000 at 5 % per annum = ( 8000 Γ 5 Γ 1 ) / 100 = rs . 400 let his additional investment at 8 % = x interest on rs . x at 8 % per annum = ( x Γ 8 Γ 1 ) / 100 = 2 x / 25 . to earn 6 % per annum for the total , interest = ( 8000 + x ) Γ 6 Γ 1 / 100 . = > 400 + 2 x / 25 = ( 8000 + x ) Γ 6 Γ 1 / 100 . = > 40000 + 8 x = ( 8000 + x ) Γ 6 . = > 40000 + 8 x = 48000 + 6 x . = > 2 x = 8000 . = > x = 4000 . answer : d | a = 8 * 1000
b = 6 - 5
c = 8 - 6
d = b / c
e = a * d
|
a ) 1720 , b ) 216 , c ) 224 , d ) 2460 , e ) 280 | e | multiply(multiply(multiply(add(3, const_1), subtract(subtract(const_10, const_1), const_1)), subtract(subtract(const_10, const_1), const_1)), subtract(subtract(subtract(const_10, const_1), const_1), const_1)) | how many 3 digit odd numbers do not use any digit more than once | "the way i solved it is a b c ( hundreds , tens , units ) c can be 1 3 5 7 9 ( any of the 5 digits ) a can be anything except ( d or 0 ) so 8 possibilitie b can be anything execpt ( a d c ) so 7 possibilities total ways are 7 * 8 * 5 = 280 ans : e" | a = 3 + 1
b = 10 - 1
c = b - 1
d = a * c
e = 10 - 1
f = e - 1
g = d * f
h = 10 - 1
i = h - 1
j = i - 1
k = g * j
|
a ) 14 β 3 , b ) 16 β 3 , c ) 15 β 3 , d ) 17 β 3 , e ) none of these | b | multiply(divide(lcm(lcm(32, 20), 24), divide(const_1, subtract(divide(const_1, 12), divide(const_1, 20)))), divide(25, add(add(add(divide(lcm(lcm(32, 20), 24), 32), divide(lcm(lcm(32, 20), 24), 20)), divide(lcm(lcm(32, 20), 24), divide(const_1, subtract(divide(const_1, 12), divide(const_1, 20))))), divide(lcm(lcm(32, 20), 24), 24)))) | a sum of 25 was paid for a work which a can do in 32 days , b in 20 days , b and c in 12 days and d in 24 days . how much did c receive if all the 4 work together ? | a ' s one day ' s work = 1 β 32 b ' s one day ' s work = 1 β 20 ( b + c ) ' s one day ' s work = 1 β 12 β΄ c ' s one day ' s work = 1 β 12 - 1 β 20 = 1 β 30 d ' s one day ' s work = 1 β 24 β΄ ( a + b + c + d ) ' s one day ' s work 1 / 32 + 1 / 20 + 1 / 30 + 1 / 24 = 75 + 120 + 80 + 100 / 2400 = 375 β 2400 = 15 β 96 = 5 β 32 β΄ out of 5 β 32 of work done 1 β 30 of the work is done by c . β out of 25 paid for the work , c will receive 1 / 30 / 5 / 32 Γ 25 , i . e , 1 β 30 Γ 32 β 5 Γ 25 , i . e , 16 β 3 answer b | a = math.lcm(32, 20)
b = math.lcm(a, 24)
c = 1 / 12
d = 1 / 20
e = c - d
f = 1 / e
g = b / f
h = math.lcm(32, 20)
i = math.lcm(h, 24)
j = i / 32
k = math.lcm(32, 20)
l = math.lcm(k, 24)
m = l / 20
n = j + m
o = math.lcm(32, 20)
p = math.lcm(o, 24)
q = 1 / 12
r = 1 / 20
s = q - r
t = 1 / s
u = p / t
v = n + u
w = math.lcm(32, 20)
x = math.lcm(w, 24)
y = x / 24
z = v + y
A = 25 / z
B = g * A
|
a ) 116636 , b ) 116406 , c ) 116647 , d ) 116291 , e ) 116675 | d | divide(multiply(add(multiply(13, const_100), 47), add(multiply(7, const_100), 77)), multiply(subtract(47, add(multiply(const_2, const_4), const_2)), subtract(47, add(multiply(const_2, const_4), const_2)))) | a room 13 m 47 cm long and 7 m 77 cm broad is to be paved with square tiles . find the least number of square tiles required to cover the floor . | "explanation : area of the room = ( 1347 x 777 ) cm 2 . size of largest square tile = h . c . f . of 1347 cm and 777 cm = 3 cm . area of 1 tile = ( 3 x 3 ) cm 2 . number of tiles required = ( 1347 Γ 777 ) / ( 3 Γ 3 ) = 116291 answer : option d" | a = 13 * 100
b = a + 47
c = 7 * 100
d = c + 77
e = b * d
f = 2 * 4
g = f + 2
h = 47 - g
i = 2 * 4
j = i + 2
k = 47 - j
l = h * k
m = e / l
|
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | a | subtract(divide(subtract(add(166, 1), 3), 2), subtract(divide(subtract(add(166, 1), 3), 2), 1)) | if a number p is prime , and 2 p + 3 = q , where q is also prime , then the decimal expansion of 1 / q will produce a decimal with q - 1 digits . if this method produces a decimal with 166 digits , what is the units digit of the product of p and q ? | "1 / 5 = 0.2 . . ( a repeating pattern one digit long ) a" | a = 166 + 1
b = a - 3
c = b / 2
d = 166 + 1
e = d - 3
f = e / 2
g = f - 1
h = c - g
|
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