euclaise/mathqa_programs huggingface dataset:
Problem: a person borrows 5000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 61 ⁄ 4 % p . a . for 2 years . find his gain in the transaction per year .
Options: a ) 112.50 , b ) 125 , c ) 150 , d ) 167.5 , e ) none of these
Rationale: "gain in 2 years = [ ( 5000 × 25 / 4 × 2 / 100 ) − ( 5000 × 4 × 2 / 100 ) ] = ( 625 – 400 ) = 225 . ∴ gain in 1 year = ( 225 ⁄ 2 ) = 112.50 answer a"
Answer: a
Python program:
a = 1 / 100
b = 5000 * a
Problem: a side of beef lost 30 percent of its weight in processing . if the side of beef weighed 500 pounds after processing , how many pounds did it weigh before processing ?
Options: a ) 714.28 , b ) 712 , c ) 714.88 , d ) 714.25 , e ) 714
Rationale: "let weight of side of beef before processing = x ( 70 / 100 ) * x = 500 = > x = ( 500 * 100 ) / 70 = 714.28 answer a"
Answer: a
Python program:
a = 500 * 100
b = 100 - 30
c = a / b
Problem: if the price of a tv is first decreased by 20 % and then increased by 40 % , then the net change in the price will be :
Options: a ) 4 % increase , b ) 12 % increase , c ) 10 % decrease , d ) 6 % increase , e ) none of these
Rationale: "explanation : solution : let the original price be rs . 100 . new final price = 140 % of ( 80 % of 100 ) = rs . 140 / 100 * 80 / 100 * 100 = rs . 112 . . ' . increase = 12 % answer : b"
Answer: b
Python program:
a = 100 - 20
b = 100 + 40 
c = a * b
d = c / 100
e = d - 100
Problem: two sets of 3 consecutive positive odd integers have exactly one integer in common . the sum of the integers in the set with greater numbers is how much greater than the sum of the integers in the other set ?
Options: a ) 4 , b ) 7 , c ) 8 , d ) 12 , e ) it can not be determined from the information given .
Rationale: a = ( 1 , 3,5 ) , sum of this = 9 b = ( 5 , 7,9 ) , sum of this = 21 , the differenct between 21 - 9 = 12 hence , 12 is the answer i . e . d
Answer: d
Python program:
a = 3 + 4 
b = a - 1
c = 3 + 2 
d = c - 3
e = 2 + 4 
f = 3 + 2 
g = f - 3
h = e - g
i = d + h 
j = b + i
Problem: a train running at 1 / 5 of its own speed reached a place in 20 hours . how much time could be saved if the train would have run at its own speed ?
Options: a ) 8 hrs , b ) 10 hrs , c ) 12 hrs , d ) 16 hrs , e ) 6 hrs
Rationale: "time taken if run its own speed = 1 / 5 * 20 = 4 hrs time saved = 20 - 4 = 16 hrs answer : d"
Answer: d
Python program:
a = 1 / 5
b = a * 20
Problem: in a certain pond , 60 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the pond ?
Options: a ) 400 , b ) 1,500 , c ) 1,250 , d ) 2,500 , e ) 10,000
Rationale: "this is a rather straight forward ratio problem . 1 . 60 fish tagged 2 . 2 out of the 50 fish caught were tagged thus 2 / 50 2 / 50 = 60 / x thus , x = 1500 think of the analogy : 2 fish is to 50 fish as 50 fish is to . . . ? you ' ve tagged 50 fish and you need to find what that comprises as a percentage of the total fish population - we have that information with the ratio of the second catch . b"
Answer: b
Python program:
a = 2 / 50
b = 60 / a
Problem: 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 ?
Options: a ) 5 , b ) 305 , c ) 365 , d ) 50 , e ) 495
Rationale: 5700 , 5710 , . . . . . . 5798 , so total 50 . hence option d .
Answer: d
Python program:
a = 4 + 7 
b = a + 7
Problem: 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 .
Options: a ) 116 , b ) 120 , c ) 130 , d ) 140 , e ) none
Rationale: 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
Answer: c
Python program:
a = 1044 / 9
b = 1044 / 9
c = b / 9
d = math.floor(c)
e = a + d
Problem: find the amount on rs . 7000 in 2 years , the rate of interest being 4 % per first year and 5 % for the second year ?
Options: a ) 5568 , b ) 6369 , c ) 5460 , d ) 7644 , e ) 6734
Rationale: "7000 * 104 / 100 * 105 / 100 = > 7644 answer : d"
Answer: d
Python program:
a = 100 + 4 
b = 7000 * a
c = b / 100
d = 100 + 5 
e = c * d
f = e / 100
Problem: 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
Options: a ) 300 , b ) 400 , c ) 600 , d ) 500 , e ) none of these
Rationale: 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 × 35 = 600 kg answer : option c
Answer: c
Python program:
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
Problem: 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 ?
Options: a ) rs . 950 , b ) rs . 1500 , c ) rs . 1000 , d ) rs . 1200 , e ) none of these
Rationale: "explanation : sp of 1 metre cloth = 8200 / 40 = rs . 205 . cp of 1 metre cloth = rs . 205 – 25 = rs . 180 cp on 40 metres = 180 x 40 = rs . 7200 profit earned on 40 metres cloth = rs . 8200 – rs . 7200 = rs . 1000 . answer : option c"
Answer: c
Python program:
a = 25 * 40
Problem: 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 ?
Options: a ) 23 , b ) 28 , c ) 27 , d ) 19 , e ) 24
Rationale: "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"
Answer: d
Python program:
a = 16 * 3
b = 64 - a
c = b + 3
Problem: 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 ?
Options: a ) 40 , b ) 65 , c ) 73 , d ) 60 , e ) 80
Rationale: "d 60 let the number be 2 + 1 / 2 [ 1 / 3 ( a / 5 ) ] = a / 15 = > 2 = a / 30 = > a = 60"
Answer: d
Python program:
a = 2 + 2 
b = 2 * a
c = 1 / b
d = c / 2
e = 2 / d
Problem: 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
Options: a ) 627 , b ) 297 , c ) 127 , d ) 150 , e ) 181
Rationale: explanation : 125 % of 120 % of a = 225 125 / 100 * 120 / 100 * a = 225 a = 225 * 2 / 3 = 150 . answer : d
Answer: d
Python program:
a = 100 * 225
b = 25 + 100 
c = a / b
d = 100 * c
e = 20 + 100 
f = d / e
Problem: if circles x and y have the same area and circle x has a circumference of 10 π , half of the radius of circle y is :
Options: a ) 16 , b ) 8 , c ) 4 , d ) 2.5 , e ) 1
Rationale: "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"
Answer: d
Python program:
a = 10 / 2
b = a / 2
Problem: how many positive integers , from 2 to 100 , inclusive , are not divisible by even integers greater than 1 ?
Options: a ) 5 , b ) 6 , c ) 8 , d ) 10 , e ) 50
Rationale: "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"
Answer: b
Python program:
a = 2 * 2
b = 2 * a
c = 2 * b
d = 2 * c
e = d * 2
f = e - 4
g = f / 10
Problem: how many terminating zeroes does 200 ! have ?
Options: a ) 40 , b ) 48 , c ) 49 , d ) 55 , e ) 64
Rationale: 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
Answer: c
Python program:
a = 4 + 1 
b = 200 / a
c = 4 + 1 
d = 4 + 1 
e = c * d
f = 200 / e
g = b + f
Problem: 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 ?
Options: a ) 65 , b ) 80 , c ) 90 , d ) 115 , e ) 120
Rationale: 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
Answer: b
Python program:
a = 40 * 200
b = 300 - 200
c = a / b
Problem: 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 ’ s selling price per item is $ 15 , how many items the manufacturer produce and sell to earn an annual profit of $ 150,000 ?
Options: a ) 2,858 , b ) 8,667 , c ) 21,429 , d ) 35,000 , e ) 70,000
Rationale: "let the items manufactured or sold bex 130000 + 11 x = 15 x - 150000 4 x = 280000 x = 70000 ans : e"
Answer: e
Python program:
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
Problem: the area of a parallelogram is 450 sq m and its altitude is twice the corresponding base . then the length of the base is ?
Options: a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 15
Rationale: "2 x * x = 450 = > x = 15 answer : e"
Answer: e
Python program:
a = 450 / 2
b = math.sqrt(a)
Problem: 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 .
Options: a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 15
Rationale: "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 )"
Answer: e
Python program:
a = 32 * 8
b = a / 10
c = b / 2
d = c - const_0_25
e = d - const_0_25
Problem: 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 :
Options: a ) 2 , b ) 4 , c ) 10 , d ) 12 , e ) 15
Rationale: "sol . speed in still water = 1 / 2 ( 15 + 5 ) kmph = 10 kmph . answer c"
Answer: c
Python program:
a = 15 + 5 
b = a / 2
Problem: find the least number which when divided by 20 , 2535 and 40 leaves remainders 14 , 1929 and 34 respectively .
Options: a ) 1394 , b ) 1294 , c ) 1194 , d ) 1094 , e ) none of them
Rationale: here , ( 20 - 14 ) = 6 , ( 25 – 19 ) = 6 , ( 35 - 29 ) = 6 and ( 40 - 34 ) = 6 . therefore , required number = ( l . c . m . of 2025 , 3540 ) – 6 = 1394 . answer is a .
Answer: a
Python program:
a = 14 / 2
b = 100 * a
c = b * 2
d = 2 * 3
e = c - d
Problem: 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 ?
Options: a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 7
Rationale: "# 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"
Answer: c
Python program:
a = 3 * 2
b = a + 4 
c = 28 / b
Problem: when average age of 22 members are 0 , how many members greater than 0 ?
Options: a ) 17 , b ) 21 , c ) 22 , d ) 24 , e ) 25
Rationale: 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 )
Answer: b
Python program:
a = 22 - 1
Problem: if 10 spiders make 5 webs in 5 days , then how many days are needed for 1 spider to make 1 web ?
Options: a ) 10 , b ) 9 , c ) 5 , d ) 15 , e ) 20
Rationale: "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 â ‡ ’ 10 ã — 1 ã — 5 = 1 ã — 5 ã — x â ‡ ’ x = 10 answer : option a"
Answer: a
Python program:
a = 1 * 10
Problem: 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 ?
Options: a ) 100 , b ) 300 , c ) 400 , d ) 3,000 , e ) 4,000
Rationale: "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 ."
Answer: c
Python program:
a = 99 - 96
b = 12 / a
c = b * 100
Problem: 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 .
Options: a ) 12.18 , b ) 10.18 , c ) 10.12 , d ) 8.1 , e ) 8.2
Rationale: 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
Answer: d
Python program:
a = 200 / 4
b = 200 + a 
c = 2000 / b
Problem: ravi and kavi start a business by investing â ‚ ¹ 4000 and â ‚ ¹ 72000 , respectively . find the ratio of their profits at the end of year .
Options: a ) 2 : 28 , b ) 5 : 28 , c ) 7 : 18 , d ) 1 : 18 , e ) 3 : 18
Rationale: "ratio of profit = ratio of investments = 4000 : 72000 = 1 : 18 answer : d"
Answer: d
Python program:
a = 4000 / 72000
Problem: 108 . triangle a ’ s base is 12 % greater than the base of triangle b , and a ’ 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 ?
Options: a ) 9 % less , b ) 0.7 % less , c ) equal to each other , d ) 1 % more , e ) 9 % more
Rationale: "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 )"
Answer: b
Python program:
a = 100 * 100
b = 100 + 12 
c = 100 - 12
d = b * c
e = a - d
f = 100 / e
Problem: 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 . ?
Options: a ) 166 , b ) 408 , c ) 400 , d ) 483 , e ) 566
Rationale: 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
Answer: b
Python program:
a = 283 * 2
b = a ** 2
c = 204 * 2
d = c ** 2
e = b - d
f = math.sqrt(e)
Problem: the triplicate ratio of 2 : 5 is ?
Options: a ) 1 : 0 , b ) 8 : 125 , c ) 1 : 7 , d ) 1 : 2 , e ) 1 : 1
Rationale: "23 : 53 = 8 : 125 answer : b"
Answer: b
Python program:
a = 2 ** 3
b = 3 ** 0
c = a / b
Problem: how many two - digit numbers that do not contain the digits 1 or 4 are there ?
Options: a ) 112 , b ) 56 , c ) 64 , d ) 49 , e ) 84
Rationale: "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"
Answer: b
Python program:
a = 10 * 10
b = 10 - 1
c = a * b
d = c * 4
Problem: 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 ?
Options: a ) 5 , b ) 48 , c ) 12 , d ) 15 , e ) 20
Rationale: 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
Answer: b
Python program:
a = 5 + 1 
b = a * 10
c = 10 / b
d = 5 + 1 
e = d * 10
f = 5 / e
g = c - f
h = 4 / g
Problem: 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
Options: a ) 70 , b ) 72 , c ) 75 , d ) 78 , e ) 80
Rationale: "income of 6 months = ( 6 × 75 ) – debt = 450 – debt income of the man for next 4 months = 4 × 60 + debt + 30 = 270 + debt ∴ income of 10 months = 720 average monthly income = 720 ÷ 10 = 72 answer b"
Answer: b
Python program:
a = 75 * 6
b = 60 * 4
c = a + b 
d = c + 30 
e = 6 + 4 
f = d / e
Problem: 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 ?
Options: a ) 1 % , b ) 4 % , c ) 6 % , d ) 8 % , e ) 9 %
Rationale: "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"
Answer: a
Python program:
a = 9 / 100
b = a + 1 
c = 2 - b
d = 1 - c
e = d * 100
Problem: 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 ?
Options: a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 60 %
Rationale: "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 ."
Answer: c
Python program:
a = 25 + 1 
b = 31 - a
c = 40 / 100
d = 25 + 1 
e = d / 100
f = c - e
g = b / f
Problem: 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 ?
Options: a ) 3 / 7 , b ) 11 / 35 , c ) 16 / 35 , d ) 31 / 70 , e ) 37 / 70
Rationale: "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 ."
Answer: b
Python program:
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
Problem: 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 ?
Options: a ) 2003 , b ) 2002 , c ) 2000 , d ) 2001 , e ) 2004
Rationale: "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"
Answer: c
Python program:
a = 8 / 10
b = a * 10000
c = 10000 - b
Problem: 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
Options: a ) 150 , b ) 200 , c ) 250 , d ) 350 , e ) none
Rationale: "sol . let the number of grils be x . then , number of boys = ( 600 - x ) . then , ( 11 3 / 4 × 600 ) ⇔ 11 x + 12 ( 600 - x ) ⇔ x = 7200 - 7050 ⇔ 150 . answer a"
Answer: a
Python program:
a = 12 * 600
b = 9 / 12
c = 11 + b 
d = c * 600
e = a - d
Problem: 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 ?
Options: a ) 20 % , b ) 64.29 % , c ) 70 % , d ) 80 % , e ) 100 %
Rationale: 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
Answer: b
Python program:
a = 100 - 50
b = 40 / 100
c = 1 + b 
d = a / c
e = 100 - d