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{ "label": [ "A", "B", "C", "D" ], "text": [ "5865863355", "5665863355", "4865863355", "4665863355" ] }
[ "A" ]
simplify 586645 * 9999
explanation : although it is a simple question , but the trick is to save time in solving this . rather than multiplying it we can do as follows : 586645 * ( 10000 - 1 ) = 5866450000 - 586645 = 5865863355 option a
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 1386", "rs . 1764", "rs . 1575", "rs . 2268" ] }
[ "B" ]
the true discount on a bill due 9 months hence at 16 % per annum is rs . 189 . the amount of the bill is :
"solution 32.5 let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 189 . ∴ x 16 x 9 / 12 x 1 / 100 = 189 or x = 1575 . ∴ p . w . = rs . 1575 . ∴ sum due = p . w . + t . d . = rs . ( 1575 + 189 ) = rs . 1764 . answer b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "25 %", "23 %", "16.67 %", "33.33 %" ] }
[ "B" ]
if the price of petrol increases by 30 , by how much must a user cut down his consumption so that his expenditure on petrol remains constant ?
"explanation : let us assume before increase the petrol will be rs . 100 . after increase it will be rs ( 100 + 30 ) i . e 130 . now , his consumption should be reduced to : - = ( 130 − 100 ) / 130 ∗ 100 . hence , the consumption should be reduced to 23 % . answer : b"
MathQA
{ "label": [ "A", "B", "C", "E" ], "text": [ "50 %", "125 %", "25 %", "30 %" ] }
[ "C" ]
the ratio 5 : 20 expressed as percent equals to
"explanation : actually it means 5 is what percent of 20 , which can be calculated as , ( 5 / 20 ) * 100 = 5 * 5 = 25 option c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "1.12", "1.2", "1.25", "1.3" ] }
[ "B" ]
if 0.75 : x : : 5 : 8 , then x is equal to :
"( x * 5 ) = ( 0.75 * 8 ) x = 6 / 5 x = 1.20 answer = b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "62 %", "73 %", "41 %", "71 %" ] }
[ "A" ]
b as a percentage of a is equal to a as a percentage of ( a + b ) . find b as a percentage of a .
explanation : according to the question : - = > b / a = a / ( a + b ) . as b is a percent of a , let us assume b = ax . then , equation ( 1 ) can be re - written as = > x = 11 + x . = > x ( 1 + x ) = 1 . = > x 2 + x + 1 = 0 . = > x = − 1 ± 5 / √ 2 thus , x = − 1 + 5 / √ 2 = > 0.62 . = > 62 % . answer : a
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "13000", "7000", "10000", "5000" ] }
[ "D" ]
a textile manufacturing firm employees 70 looms . it makes fabrics for a branded company . the aggregate sales value of the output of the 70 looms is rs 00000 and the monthly manufacturing expenses is rs 1 , 50000 . assume that each loom contributes equally to the sales and manufacturing expenses are evenly spread over the number of looms . monthly establishment charges are rs 75000 . if one loom breaks down and remains idle for one month , the decrease in profit is :
explanation : profit = 5 , 00,000 â ˆ ’ ( 1 , 50,000 + 75,000 ) = rs . 2 , 75,000 . since , such loom contributes equally to sales and manufacturing expenses . but the monthly charges are fixed at rs 75,000 . if one loan breaks down sales and expenses will decrease . new profit : - = > 500000 ã — ( 69 / 70 ) â ˆ ’ 150000 ã — ( 69 / 70 ) â ˆ ’ 75000 . = > rs 2 , 70,000 . decrease in profit = > 2 , 75,000 â ˆ ’ 2 , 70,000 = > rs . 5,000 . answer : d
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 11.81", "rs . 12", "rs . 18.94", "rs . 12.31" ] }
[ "C" ]
a fruit seller sells mangoes at the rate of rs . 14 per kg and thereby loses 15 % . at what price per kg , he should have sold them to make a profit of 15 % ?
"solution 85 : 14 = 115 : x x = ( 14 ã — 115 / 85 ) = rs . 18.94 hence , s . p per kg = rs . 18.94 answer c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "6 hours", "5 hours", "7 hours", "8 hours" ] }
[ "C" ]
a truck covers a distance of 392 km at a certain speed in 8 hours . how much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 70 km more than that travelled by the truck ?
"explanation : speed of the truck = distance / time = 392 / 8 = 49 kmph now , speed of car = ( speed of truck + 18 ) kmph = ( 48 + 18 ) = 66 kmph distance travelled by car = 392 + 70 = 462 km time taken by car = distance / speed = 462 / 66 = 7 hours . answer – c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 435", "rs . 350", "rs . 275", "rs . 425" ] }
[ "A" ]
the mean daily profit made by a shopkeeper in a month of 30 days was rs . 350 . if the mean profit for the first fifteen days was rs . 265 , then the mean profit for the last 15 days would be
average would be : 350 = ( 265 + x ) / 2 on solving , x = 435 . answer : a
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "427", "859", "869", "856" ] }
[ "D" ]
the least number which when increased by 8 each divisible by each one of 24 , 32 , 36 and 54 is :
"solution required number = ( l . c . m . of 24 , 32 , 36 , 54 ) - 8 = 864 - 8 = 856 . answer d"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "11", "17", "13", "20" ] }
[ "D" ]
when 242 is divided by a certain divisor the remainder obtained is 11 . when 698 is divided by the same divisor the remainder obtained is 18 . however , when the sum of the two numbers 242 and 698 is divided by the divisor , the remainder obtained is 9 . what is the value of the divisor ?
"let that divisor be x since remainder is 11 or 18 it means divisor is greater than 18 . now 242 - 11 = 231 = kx ( k is an integer and 234 is divisble by x ) similarly 698 - 18 = 680 = lx ( l is an integer and 689 is divisible by x ) adding both 698 and 242 = ( 231 + 680 ) + 11 + 18 = x ( k + l ) + 29 when we divide this number by x then remainder will be equal to remainder of ( 29 divided by x ) = 9 hence x = 29 - 9 = 20 hence d"
MathQA
{ "label": [ "A", "B", "C", "E" ], "text": [ "3", "4", "5", "6" ] }
[ "B" ]
the smallest value of n , for which 2 n + 1 is not a prime number , is
"solution ( 2 × 1 + 1 ) = 3 . ( 2 × 2 + 1 ) = 5 . ( 2 × 3 + 1 ) = 7 . ( 2 × 4 + 1 ) = 9 . which is not prime , n = 4 . answer b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "1865113", "1775123", "1765013", "1675123" ] }
[ "C" ]
( 4300631 ) - ? = 2535618
"let 4300631 - x = 2535618 then x = 4300631 - 2535618 = 1765013 answer is c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 170", "rs . 140", "rs . 1.70", "rs . 4.25" ] }
[ "B" ]
if 0.5 % of a = 70 paise , then the value of a is ?
"answer ∵ 0.5 / 100 of a = 70 / 100 ∴ a = rs . ( 70 / 0.5 ) = rs . 140 correct option : b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs 200", "rs 220", "rs 260", "rs 280" ] }
[ "A" ]
the sale price of a trolley bag including the sale tax is rs . 280 . the rate of sale tax is 12 % . if the shopkeeper has made a profit of 25 % , the cost price of the trolley bag is :
"explanation : 112 % of s . p . = 280 s . p . = rs . ( 280 x 100 / 112 ) = rs . 250 . c . p . = rs ( 100 / 125 x 250 ) = rs 200 answer : a"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "4.37 %", "2 %", "6 %", "8.75 %" ] }
[ "B" ]
the population of a town increased from 1 , 25,000 to 1 , 50,000 in a decade . the average percent increase of population per year is :
"solution increase in 10 year = ( 150000 - 125000 ) = 25000 . increase % = ( 25000 / 125000 x 100 ) % = 20 % â ˆ ´ required average = ( 20 / 10 ) % = 2 % answer b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "85", "93", "83", "72" ] }
[ "B" ]
the sum of digits of a two digit number is 12 , the difference between the digits is 6 . find the number
"description : = > x + y = 12 , x - y = 6 adding these 2 x = 18 = > x = 9 , y = 3 . thus the number is 93 answer b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 72", "rs . 36", "rs . 55", "rs . 50" ] }
[ "C" ]
the banker ' s gain on a bill due due 1 year hence at 12 % per annum is rs . 6.6 . the true discount is
solution t . d = [ b . g x 100 / r x t ] = rs . ( 6.6 x 100 / 12 x 1 ) = rs . 55 . answer c
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "46 kg", "48 kg", "70 kg", "72 kg" ] }
[ "A" ]
the average weight of 8 students decreases by 5 kg when one of them weighing 86 kg is replaced by a new student . the weight of the student is
"explanation : let the weight of student be x kg . given , difference in average weight = 5 kg = > ( 86 - x ) / 8 = 5 = > x = 46 answer : a"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "28800", "24000", "24936", "25640" ] }
[ "A" ]
80 % of the population of a village is 23040 . the total population of the village is ?
"answer ∵ 80 % of p = 23040 ∴ p = ( 23040 x 100 ) / 80 = 28800 correct option : a"
MathQA
{ "label": [ "A", "C", "D", "E" ], "text": [ "4", "2", "6", "1" ] }
[ "E" ]
if p is a prime number greater than 5 , what is the remainder when p ^ 2 is divided by 6 .
take square of any prime number remainder will be 1 ans e
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "200", "230", "140", "170" ] }
[ "D" ]
in the annual cultural programme of indra prastha there was this math quiz going on . yudhisthir faced this last question that will bring the champion ' s trophy . what will be thesolution for the given problem ? the function f is defined is defined asf ( 2 x + 3 ) = ( x - 3 ) ( x + 4 ) what is f ( 29 ) ?
explanation : f ( 2 x + 3 ) = ( x - 3 ) ( x + 4 ) put x = 13 f ( 2 * 13 + 3 ) = ( 13 - 3 ) * ( 13 + 4 ) = 10 * 17 f ( 29 ) = 170 answer : d
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "4 : 3", "8 : 7", "4 : 1", "6 : 5" ] }
[ "B" ]
an order was placed for the supply of a carper whose length and breadth were in the ratio of 3 : 2 . subsequently , the dimensions of the carpet were altered such that its length and breadth were in the ratio 7 : 3 but were was no change in its parameter . find the ratio of the areas of the carpets in both the cases .
"explanation : let the length and breadth of the carpet in the first case be 3 x units and 2 x units respectively . let the dimensions of the carpet in the second case be 7 y , 3 y units respectively . from the data , . 2 ( 3 x + 2 x ) = 2 ( 7 y + 3 y ) = > 5 x = 10 y = > x = 2 y required ratio of the areas of the carpet in both the cases = 3 x * 2 x : 7 y : 3 y = 6 x 2 : 21 y 2 = 6 * ( 2 y ) 2 : 21 y 2 = 6 * 4 y 2 : 21 y 2 = 8 : 7 answer is b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "18 %", "20 %", "17 %", "18 %" ] }
[ "C" ]
a shopkeeper labeled the price of his articles so as to earn a profit of 30 % on the cost price . he then sold the articles by offering a discount of 10 % on the labeled price . what is the actual percent profit earned in the deal ?
explanation : let the cp of the article = rs . 100 . then labeled price = rs . 130 . sp = rs . 130 - 10 % of 130 = rs . 130 - 13 = rs . 117 . gain = rs . 117 – rs . 100 = rs . 17 therefore , gain / profit percent = 17 % . answer : option c
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "100 m", "120 m", "140 m", "160 m" ] }
[ "C" ]
a train covers a distance of 12 km in 10 minutes . if it takes 7 seconds to pass a telegraph post , then the length of the train is
"explanation : speed = 12 / 10 x 60 km / hr = 72 x 5 / 18 m / sec = 20 m / sec . length of the train = ( speed x time ) = ( 20 x 7 ) m = 140 m answer : option c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "296', '", "252', '", "344', '", "388', '" ] }
[ "C" ]
the sum of four consecutive even numbers is 36 . find the sum of the squares of these numbers ?
let the four numbers be x , x + 2 , x + 4 and x + 6 . = > x + x + 2 + x + 4 + x + 6 = 36 = > 4 x + 12 = 36 = > x = 6 the numbers are 6 , 8 , 10 and 12 . sum of their squares = 62 + 82 + 102 + 122 = 36 + 64 + 100 + 144 = 344 . answer : c
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "48 hours", "51 hours", "36 hours", "56 hours" ] }
[ "D" ]
speed of a boat in standing water is 8 kmph and the speed of the stream is 2 kmph . a man rows to place at a distance of 210 km and comes back to the starting point . the total time taken by him is :
"sol . speed upstream = 6 kmph ; speed downstream = 10 kmph . ∴ total time taken = [ 210 / 6 + 210 / 10 ] hours = 56 hours . answer d"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 880", "rs . 890", "rs . 1200", "rs . 900" ] }
[ "C" ]
if the compound interest on a certain sum of money for 4 years at 10 % per annum be rs . 993 , what would be the simple interest ?
"let p = principal a - amount we have a = p ( 1 + r / 100 ) 3 and ci = a - p atq 993 = p ( 1 + r / 100 ) 3 - p ? p = 3000 / - now si @ 10 % on 3000 / - for 4 yrs = ( 3000 x 10 x 4 ) / 100 = 1200 / - answer : c ."
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "3", "2", "1 / 2", "1 / 3" ] }
[ "D" ]
two brothers took the gmat exam , the higher score is u and the lower one is v . if the difference between the two scores is equal to their average , what is the value of v / u ?
answer is d : 1 / 3 u - v = ( u + v ) / 2 solving for v / u = 1 / 3 d
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "2.09 %", "20.9 %", "209 %", "0.209 %" ] }
[ "C" ]
2.09 can be expressed in terms of percentage as
"explanation : while calculation in terms of percentage we need to multiply by 100 , so 2.09 * 100 = 209 answer : option c"
MathQA
{ "label": [ "A", "B", "D", "E" ], "text": [ "3", "1 / 2", "- 1 / 2", "- 1" ] }
[ "A" ]
if | x | = 4 x - 1 , then x = ?
"answer : a approach : substituted option a i . e x = 1 . inequality satisfied ."
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "12.50 %", "13.50 %", "5 %", "14.50 %" ] }
[ "C" ]
an uneducated retailer marks all his goods at 40 % above the cost price and thinking that he will still make 25 % profit , offers a discount of 25 % on the marked price . what is his actual profit on the sales ?
sol . let c . p . = rs . 100 . then , marked price = rs . 140 . s . p . = 75 % of rs . 140 = rs . 105 . ∴ gain % = 5 % . answer c
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "800", "3200", "900", "1600" ] }
[ "B" ]
find the value of 4 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ]
"answer 4 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ] = 4 x [ ( 36 x 48 x 250 ) / ( 12 x 9 x 5 ) ] = 4 x 4 x 4 x 50 = 3200 correct option : b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "16000", "18000", "20000", "30000" ] }
[ "D" ]
a courtyard is 18 meter long and 12 meter board is to be paved with bricks of dimensions 12 cm by 6 cm . the total number of bricks required is :
"explanation : number of bricks = courtyard area / 1 brick area = ( 1800 ã — 1200 / 12 ã — 6 ) = 30000 option d"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "9 %", "9.27 %", "22.50 %", "12 %" ] }
[ "C" ]
the population of a city increases by 8 % per year but due to migration it decrease by 1 % per years . what will be the percentage increase in population in 3 years ?
"actual increase in population = 7 % let , earlier population = 100 then the population after 3 years = 100 ( 1 + 7 / 100 ) ^ 3 = 122.5043 ∴ required percentage = 22.50 % answer : c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "546", "674", "672", "960" ] }
[ "A" ]
in an office in singapore there are 60 % female employees . 50 % of all the male employees are computer literate . if there are total 62 % employees computer literate out of total 1300 employees , then the no . of female employees who are computer literate ?
"solution : total employees , = 1300 female employees , 60 % of 1300 . = ( 60 * 1300 ) / 100 = 780 . then male employees , = 520 50 % of male are computer literate , = 260 male computer literate . 62 % of total employees are computer literate , = ( 62 * 1300 ) / 100 = 806 computer literate . thus , female computer literate = 806 - 260 = 546 . answer : option a"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "12 kg", "120 kg", "72 kg", "96 kg" ] }
[ "B" ]
a boat having a length 4 m and breadth 3 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of the man is :
"explanation : volume of water displaced = ( 4 x 3 x 0.01 ) m 3 = 0.12 m 3 . ∴ mass of man = volume of water displaced x density of water = ( 0.12 x 1000 ) kg = 120 kg . answer : b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "20 kg", "25 kg", "30 kg", "35 kg" ] }
[ "B" ]
when the price of sugar was increased by 32 % , a family reduced its consumption in such a way that the expenditure on sugar was only 10 % more than before . if 30 kg were consumed per month before , find the new monthly consumption .
since , expenditure = price × consumption ∴ 110 % of 30 = 132 ⁄ 100 × new consumption ⇒ 110 ⁄ 100 × 30 = 132 ⁄ 100 × new consumption ⇒ new consumption = 25 kg answer b
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 2000", "rs . 10,500", "rs . 15,500", "rs . 22,500" ] }
[ "D" ]
a man took a loan from a bank at the rate of 8 % p . a . simple interest . after 3 years he had to pay rs . 5400 interest only for the period . the principal amount borrowed by him was :
solution principal = rs . ( 100 x 5400 / 8 x 3 ) = rs . 22,500 . answer d
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "12 kg", "60 kg", "72 kg", "96 kg" ] }
[ "B" ]
a boat having a length 3 m and breadth 2 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of man is :
"solution volume of water displaced = ( 3 × 2 × 0.01 ) m 3 = 0.06 m 3 . ∴ mass of man = volume of water displaced × density of water = ( 0.06 × 100 ) kg = 60 kg . answer b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "78 kg", "60 kg", "64 kg", "70 kg" ] }
[ "A" ]
zinc and copper are melted together in the ratio 9 : 11 . what is the weight of melted mixture , if 35.1 kg of zinc has been consumed in it ?
"sol . for 9 kg zinc , mixture melted = ( 9 + 11 ) kg . for 35.1 kg zinc , mixture , melted = [ 20 / 9 x 35.1 ] kg = 78 kg . answer a"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "6", "12", "24", "120" ] }
[ "C" ]
the largest natural number which exactly divides the product of any 4 consecutive natural numbers is
sol . required number = 1 × 2 × 3 × 4 = 24 . answer c
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "3327", "3237", "3927", "2337" ] }
[ "C" ]
9873 + x = 13800 , then x is ?
"answer x = 13800 - 9873 = 3927 option : c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "45 ( 4 / 11 ) %", "45 %", "53 ( 11 / 13 ) %", "44 ( 5 / 11 ) %" ] }
[ "C" ]
a batsman scored 130 runs which included 3 boundaries and 8 sixes . what percent of his total score did he make by running between the wickets ?
"explanation : total runs scored = 130 total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60 total runs scored by running between the wickets = 130 - 60 = 70 required % = ( 70 / 130 ) × 100 = 700 / 13 = 53 ( 11 / 13 ) % answer : option c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "50", "60", "80", "320" ] }
[ "A" ]
a number exceeds 20 % of itself by 40 . the number is
"explanation : let the answer be ' a ' then a − 20 / 100 a = 40 ⇒ 5 a − a = 200 ⇒ a = 50 ⇒ a = 50 correct option : a"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "120,50", "180,60", "180,90", "120,40" ] }
[ "A" ]
a third of arun â € ™ s marks in mathematics exceed a half of his marks in english by 15 . if he got 170 marks in two subjects together how many marks did he got in english ?
let arun â € ™ s marks in mathematics and english be x and y then ( 1 / 3 ) x - ( 1 / 2 ) y = 15 2 x - 3 y = 90 â € ¦ â € ¦ > ( 1 ) x + y = 170 â € ¦ â € ¦ . > ( 2 ) solving ( 1 ) and ( 2 ) x = 120 and y = 50 answer is a .
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "178", "176", "175", "96" ] }
[ "A" ]
a train passes a man standing on a platform in 8 seconds and also crosses the platform which is 267 metres long in 20 seconds . the length of the train ( in metres ) is :
explanation : let the length of train be l m . acc . to question ( 267 + l ) / 20 = l / 8 2136 + 8 l = 20 l l = 2136 / 12 = 178 m answer a
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "160", "180", "200", "240" ] }
[ "D" ]
a number whose fifth part increased by 6 is equal to its fourth part diminished by 6 is ?
"answer let the number be n . then , ( n / 5 ) + 6 = ( n / 4 ) - 6 â ‡ ’ ( n / 4 ) - ( n / 5 ) = 12 â ‡ ’ ( 5 n - 4 n ) / 20 = 12 â ˆ ´ n = 240 option : d"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "1 : 4", "1 : 25", "1 : 52", "1 : 522" ] }
[ "A" ]
a cube of edge 6 cm is cut into cubes each of edge 3 cm . the ratio of the total surface area of one of the small cubes to that of the large cube is equal to :
"sol . required ratio = 6 * 3 * 3 / 6 * 6 * 6 = 1 / 4 = 1 : 4 . answer a"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 120", "rs . 160", "rs . 240", "rs . 330" ] }
[ "D" ]
a sum of rs . 1870 has been divided among a , b and c such that a gets of what b gets and b gets of what c gets . b ’ s share is :
"explanation let c ’ s share = rs . x then , b ’ s share = rs . x / 4 , a ’ s share = rs . ( 2 / 3 x x / 4 ) = rs . x / 6 = x / 6 + x / 4 + x = 1870 = > 17 x / 12 = 1870 = > 1870 x 12 / 17 = rs . 1320 hence , b ’ s share = rs . ( 1320 / 4 ) = rs . 330 . answer d"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "0.1484", "14.84", "1.484", "2.762" ] }
[ "C" ]
by how much is 12 % of 24.2 more than 10 % of 14.2 ?
"answer required difference = ( 12 x 24.2 ) / 100 - ( 10 x 14.2 ) / 100 = 2.904 - 1.42 = 1.484 . correct option : c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "60 kg", "63 kg", "50 kg", "77 kg" ] }
[ "B" ]
how many kilograms of sugar costing rs . 9 per kg must be mixed with 27 kg of sugar costing rs . 7 per kg so that there may be gain of 10 % by selling the mixture at rs . 9.24 per kg ?
"explanation : let the rate of second quality be rs x per kg . step 1 : s . p of 1 kg of mixture = rs . 9.24 gain = 10 % c . p of 1 kg of mixture = [ 100 / ( 100 + 10 ) × 9.24 ] . = > rs . 8.40 . thus , the mean price = rs . 8.40 . step 2 : c . p of 1 kg of sugar of 1 st kind = 900 p c . p of 1 kg of sugar of 2 nd kind = 700 p mean price = 840 p by the rule of alligation , we have : c . p . of 1 kg of c . p . of 1 kg of sugar of 1 st sugar of 2 nd kind ( 900 p ) kind ( 700 p ) \ / mean price ( 840 p ) / \ 840 - 700 : 900 - 840 ( 140 ) ( 60 ) = > required ratio = 140 : 60 = 7 : 3 . step 3 : let , x kg of sugar of 1 st kind be mixed with 27 kg of 2 nd kind , then = > 7 : 3 = x : 27 . = > 7 / 3 = x / 27 . = > x = ( 7 / 3 ) x 27 . = > x = 63 . answer : b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 692.80", "rs . 820", "rs . 1170", "rs . 1385" ] }
[ "C" ]
average monthly income of a family of 4 earning members was rs . 735 . one of the earning members died and therefore , the average income came down to rs 590 . the income of the deceased was ?
"answer income of the deceased = total income of 4 members - total income of remaining 3 members . = 735 x 4 - 590 x 3 rs . = 1170 rs . correct option : c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "144", "288", "36", "256" ] }
[ "C" ]
in how many ways 3 boys and 3 girls can be seated in a row so that they are alternate .
"solution : let the arrangement be , b g b g b g b 34 boys can be seated in 3 ! ways . girl can be seated in 3 ! ways . required number of ways , = 3 ! * 3 ! = 36 . answer : option c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "95", "99", "26", "73" ] }
[ "B" ]
what two - digit number is less than the sum of the square of its digits by 13 and exceeds their doubled product by 5 ?
let the digits be x and y . the number would be 10 x + y . we are given that 2 xy + 5 = 10 x + y = x ^ 2 y ^ 2 - 13 thus 2 xy + 5 = x ^ 2 + y ^ 2 - 13 x ^ 2 + y ^ 2 - 2 xy = 16 ( x - y ) ^ 2 = 16 ( x - y ) = 4 or - 4 substituting the values of ( x - y ) in the equation 2 xy + 5 = 10 x + y x comes out to be 1 or 9 . . . thus the two numbers can be 15 or 99 thus the answer is b
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "20 %", "25 %", "55 %", "65 %" ] }
[ "A" ]
the ratio 4 : 20 expressed as percent equals to
explanation : actually it means 4 is what percent of 20 , which can be calculated as , ( 4 / 20 ) * 100 = 4 * 5 = 20 answer : option a
MathQA
{ "label": [ "B", "C", "D", "E" ], "text": [ "$ 81.90", "$ 8190.03", "$ 819.00", "$ 8.19" ] }
[ "D" ]
jack takes a loan of $ 120000 with 12 % annual interest : the interest is paid once , at the end of the year . jill takes a loan of $ 120000 with 12 % annual interest , compounding monthly at the end of each month . at the end of one full year , compared to jack ' s loan interest , approximately how much more does jill have to repay ?
jack ' s interest = $ 120,000 * 0.12 = $ 14400 or $ 1,200 each month . jills ’ s interest , 12 % / 12 = 1 % each month : for the 1 st month = $ 120,000 * 0.01 = $ 1,200 ; for the 2 nd month = $ 1,200 + 1 % of 1,200 = $ 1,212 , so we would have interest earned on interest ( very small amount ) ; for the 3 rd month = $ 1,212 + 1 % of 1,212 = ~ $ 1,224 ; for the 4 th month = $ 1,224 + 1 % of 1,224 = ~ $ 1,236 ; . . . for the 12 th month = $ 1,320 + 1 % of 1,320 = ~ $ 1,332 . the difference between jack ' s interest and jill ’ s interest = ~ ( 12 + 24 + . . . + 132 ) = $ 792 . answer : d .
MathQA
{ "label": [ "A", "B", "C", "E" ], "text": [ "700", "900", "705", "506" ] }
[ "B" ]
evaluate 45 / . 05
"explanation : 45 / . 05 = 4500 / 5 = 900 option b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "2000", "2573", "1600", "2950" ] }
[ "C" ]
solve for x and check : - 200 x = 1600
"solution : dividing each side by - 200 , we obtain ( - 200 x / - 200 ) = ( 1600 / - 200 ) therefore : x = - 8 check : - 200 x = 1600 ( - 200 * - 8 ) = 1600 1600 = 1600 answer : c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "2 / 5", "3 / 5", "1 / 6", "1" ] }
[ "D" ]
what is difference between biggest and smallest fraction among 2 / 3 , 3 / 4 , 4 / 5 and 5 / 3
"explanation : 2 / 3 = . 66 , 3 / 4 = . 75 , 4 / 5 = . 8 and 5 / 3 = 1.66 so biggest is 5 / 3 and smallest is 2 / 3 their difference is 5 / 3 - 2 / 3 = 3 / 3 = 1 option d"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "12.4", "14", "16", "18.6" ] }
[ "C" ]
the average of 10 numbers is calculated as 15 . it is discovered later on that while calculating the average one number , namely 36 was wrongly read as 26 . the correct average is :
explanation : sum of numbers = ( 10 × × 15 - 26 + 36 ) = 160 correct average = 160 / 10 = 16 correct option : c
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "600", "750", "1000", "1250" ] }
[ "D" ]
a man walking at the rate of 5 km / hr crosses a bridge in 15 minutes . the length of the bridge ( in metres ) is
"explanation : speed = ( 5 × 5 / 18 ) m / sec = 25 / 18 m / sec . distance covered in 15 minutes = ( 25 / 18 × 15 × 60 ) m = 1250 m . answer : d"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "1 : 5", "9 : 25", "5 : 1", "25 : 9" ] }
[ "C" ]
two interconnected , circular gears travel at the same circumferential rate . if gear a has a diameter of 10 centimeters and gear b has a diameter of 50 centimeters , what is the ratio of the number of revolutions that gear a makes per minute to the number of revolutions that gear b makes per minute ?
"same circumferential rate means that a point on both the gears would take same time to come back to the same position again . hence in other words , time taken by the point to cover the circumference of gear a = time take by point to cover the circumference of gear b time a = 2 * pi * 25 / speed a time b = 2 * pi * 5 / speed b since the times are same , 50 pi / speed a = 10 pi / speed b speeda / speed b = 50 pi / 30 pi = 5 / 1 correct option : c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 80", "rs . 86", "rs . 90", "rs . 95" ] }
[ "B" ]
a trader sells 45 meters of cloth for rs . 4500 at the profit of rs . 14 per metre of cloth . what is the cost price of one metre of cloth ?
"sp of 1 m of cloth = 4500 / 45 = rs . 100 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 105 - rs . 14 = rs . 86 . answer : b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "5 : 2", "7 : 3", "9 : 2", "13 : 4" ] }
[ "B" ]
the age of father 10 years ago was thirce the age of his son . 10 years hence , father ’ s age will be twice that of his son . the ration of their present ages is :
solution let the ages of father and son 10 year ago be 3 x and x years respectively . then , ( 3 x + 10 ) + 10 = 2 [ ( x + 10 ) + 10 ⇔ 3 x + 20 = 2 x + 40 ⇔ x = 20 . ∴ required ratio = ( 3 x + 10 ) : ( x + 10 ) = 70 : 30 : 7 : 3 . answer b
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "100', '", "120', '", "150', '", "180', '" ] }
[ "B" ]
one side of a rectangular field is 15 m and one of its diagonal is 17 m . find the area of the field .
solution other side = √ ( 17 ) 2 - ( 15 ) 2 = √ 289 - 225 = √ 64 = 8 m . ∴ area = ( 15 x 8 ) m 2 = 120 m 2 . answer b
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs 840", "rs 1320", "rs 1620", "rs 1890" ] }
[ "D" ]
if 4 men working 10 hours a day earn rs . 1400 per week , then 9 men working 6 hours a day will earn how much per week ?
"explanation : ( men 4 : 9 ) : ( hrs / day 10 : 6 ) : : 1400 : x hence 4 * 10 * x = 9 * 6 * 1400 or x = 9 * 6 * 1400 / 4 * 10 = 1890 answer : d"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "350", "370", "390", "430" ] }
[ "B" ]
average expenditure of a person for the first 3 days of a week is rs . 330 and for the next 4 days is rs . 420 . average expenditure of the man for the whole week is :
"explanation : assumed mean = rs . 330 total excess than assumed mean = 4 × ( rs . 420 - rs . 350 ) = rs . 280 therefore , increase in average expenditure = rs . 280 / 7 = rs . 40 therefore , average expenditure for 7 days = rs . 330 + rs . 40 = rs . 370 correct option : b"
MathQA
{ "label": [ "A", "B", "D", "E" ], "text": [ "32", "37", "43", "50" ] }
[ "E" ]
set x consists of 10 integers and has median of 30 and a range of 30 . what is the value of the greatest possible integer that can be present in the set ?
"note that both median and range do not restrict too many numbers in the set . range is only concerned with the smallest and greatest . median only cares about the middle . quick check of each option starting from the largest : ( e ) 50 range of 20 means the smallest integer will be 30 . so 20 can not lie in between and hence can not be the median . ( d ) 43 range of 20 means the smallest integer will be 23 . so 20 can not lie in between and hence can not be the median . ( c ) 40 range of 20 means the smallest integer will be 20 . 20 can lie in between such as : 20 , 20 , 20 , 20 , 20 , 20 , 20 , 20 , 20 , 40 , 50 this is possible . hence it is the greatest such number . answer ( e )"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "15 %", "20 %", "23 %", "30 %" ] }
[ "C" ]
if the price of sugar rises from rs . 10 per kg to rs . 13 per kg , a person , to have no increase in the expenditure on sugar , will have to reduce his consumption of sugar by
"sol . let the original consumption = 100 kg and new consumption = x kg . so , 100 x 10 = x × 13 = x = 77 kg . ∴ reduction in consumption = 23 % . answer c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "80 kmph", "69 kmph", "70 kmph", "90 kmph" ] }
[ "B" ]
a train travels 290 km in 4.5 hours and 400 km in 5.5 hours . find the average speed of train .
"as we know that speed = distance / time for average speed = total distance / total time taken thus , total distance = 290 + 400 = 690 km thus , total speed = 10 hrs or , average speed = 690 / 10 or , 69 kmph . answer : b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "1584', '", "1854', '", "1458', '", "1485', '" ] }
[ "A" ]
the radius of a cylinder is 12 m , height 21 m . the lateral surface area of the cylinder is :
lateral surface area = 2 π rh = 2 × 22 / 7 × 12 × 21 = 44 × 36 = 1584 m ( power 2 ) answer is a .
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "10 %", "50 %", "25 %", "17.6 %" ] }
[ "D" ]
a retailer buys a radio for rs 225 . his overhead expenses are rs 30 . he sellis the radio for rs 300 . the profit percent of the retailer is
"explanation : cost price = ( 225 + 30 ) = 255 sell price = 300 gain = ( 45 / 255 ) * 100 = 17.6 % . answer : d"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "41", "910", "1001", "1911" ] }
[ "A" ]
the maximum numbers of students among them 451 pens and 410 toys can be distributed in such a way that each student gets the same number of pens and same number of toys is
"olution required number of students . = h . c . f of 451 and 410 . â € ¹ = â € º 41 . answer a"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "120 m .", "80 m .", "150 m .", "100 m ." ] }
[ "A" ]
a runs twice as fast as b and gives b a start of 60 m . how long should the racecourse be so that a and b might reach in the same time ?
ratio of speeds of a and b is 2 : 1 b is 60 m away from a but we know that a covers 1 meter ( 2 - 1 ) more in every second than b the time taken for a to cover 60 m is 60 / 1 = 60 m so the total time taken by a and b to reach = 2 * 60 = 120 m answer : a
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "20 , 35,45", "28 , 49,63", "16 , 28,36", "16 , 28,46" ] }
[ "B" ]
the present ages of 3 persons are in the proportion of 4 : 7 : 9 . 8 years ago , the sum of their ages was 116 . find their present ages .
let the present ages of three persons be 4 k , 7 k and 9 k respectively . ( 4 k - 8 ) + ( 7 k - 8 ) + ( 9 k - 8 ) = 116 20 k = 140 k = 7 therefore , then present ages are 28 , 49,63 . answer : b
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "81 min", "108 min", "144 min", "192 min" ] }
[ "C" ]
one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill the tank in 86 minutes , then the slower pipe alone will be able to fill the tank in
"solution let the slower pipe alone fill the tank in x minutes . then , faster pipes will fill it in x / 3 minutes . therefore , 1 / x + 3 / x = 1 / 36 ‹ = › 4 / x = 1 / 36 ‹ = › x = 144 min . answer c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 600", "rs . 500", "rs . 400", "rs . 300" ] }
[ "A" ]
the banker ' s gain of a certain sum due 2 years hence at 10 % per annum is rs . 24 . what is the present worth ?
"explanation : t = 2 years r = 10 % td = ( bg × 100 ) / tr = ( 24 × 100 ) / ( 2 × 10 ) = 12 × 10 = rs . 120 td = ( pw × tr ) / 100 ⇒ 120 = ( pw × 2 × 10 ) / 100 ⇒ 1200 = pw × 2 pw = 1200 / 2 = rs . 600 answer : option a"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "960", "1060", "1,200", "1020" ] }
[ "D" ]
the profit earned by selling an article for 912 is equal to the loss incurred when the same article is sold for 448 . what should be the sale price of the article for making 50 per cent profit ?
"let the profit or loss be x and 912 – x = 448 + x or , x = 464 ⁄ 2 = 232 \ cost price of the article = 912 – x = 448 + x = 680 \ sp of the article = 680 × 150 ⁄ 100 = 1020 answer d"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "25", "60", "90", "140" ] }
[ "B" ]
if the average ( arithmetic mean ) of a and b is 30 and the average of b and c is 60 , what is the value of c − a ?
- ( a + b = 60 ) b + c = 120 c - a = 60 b . 60
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "1.9732", "1.0025", "1.5693", "1.0266" ] }
[ "A" ]
given that 268 x 74 = 19732 , find the value of 2.68 x . 74 .
"solution sum of decimals places = ( 2 + 2 ) = 4 . therefore , = 2.68 × . 74 = 1.9732 answer a"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "10 %", "11 %", "12 %", "10.5 %" ] }
[ "B" ]
on a sum of money , the simple interest for 2 years is rs . 660 , while the compound interest is rs . 696.30 , the rate of interest being the same in both the cases . the rate of interest is
"solution difference in c . i and s . i for 2 years = rs ( 696.30 - 660 ) = rs . 36.30 . s . i for one years = rs 330 . s . i on rs . 330 for 1 year = rs . 36.30 rate = ( 100 x 36.30 / 330 x 1 ) % = 11 % . answer b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "5 hours", "4 hours", "3 hours", "2 hours" ] }
[ "C" ]
a boat can travel with a speed of 22 km / hr in still water . if the speed of the stream is 5 km / hr , find the time taken by the boat to go 81 km downstream
"explanation : speed of the boat in still water = 22 km / hr speed of the stream = 5 km / hr speed downstream = ( 22 + 5 ) = 27 km / hr distance travelled downstream = 81 km time taken = distance / speed = 81 / 27 = 3 hours . answer : option c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "22.2 %", "36 %", "80 %", "122.2 %" ] }
[ "D" ]
the cost price of an article is 36 % of the marked price . calculate the gain percent after allowing a discount of 20 % .
"sol . let marked price = rs . 100 . then , c . p . = rs . 36 . s . p = rs . 80 . â ˆ ´ gain % = [ 44 / 36 * 100 ] % = 122.2 % . answer d"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "18 %", "21 %", "25.9 %", "19 %" ] }
[ "C" ]
mahesh marks an article 15 % above the cost price of rs . 540 . what must be his discount percentage if he sells it at rs . 460 ?
cp = rs . 540 , mp = 540 + 15 % of 540 = rs . 621 sp = rs . 460 , discount = 621 - 460 = 161 discount % = 161 / 621 * 100 = 25.9 % answer : c
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "1 / 5", "1 / 6", "1 / 7", "3 / 10" ] }
[ "D" ]
a can finish a work in 10 days and b can do same work in half the time taken by a . then working together , what part of same work they can finish in a day ?
"explanation : please note in this question , we need to answer part of work for a day rather than complete work . it was worth mentioning here because many do mistake at this point in hurry to solve the question so lets solve now , a ' s 1 day work = 1 / 10 b ' s 1 day work = 1 / 5 [ because b take half the time than a ] ( a + b ) ' s one day work = ( 1 / 10 + 1 / 5 ) = 3 / 10 so in one day 3 / 10 work will be done answer : d"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "14", "42", "28", "12" ] }
[ "B" ]
two sets of 6 consecutive positive 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 ?
"a = ( 1,2 , 3,4 , 5,6 ) , sum of this = 21 b = ( 6 , 7,8 , 9,10 , 11,12 ) , sum of this = 63 , the differenct between 63 - 21 = 42 hence , 42 is the answer i . e . b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "56.25", "60", "52.5", "5.25" ] }
[ "A" ]
what is the sum of the first 15 terms of an a . p whose 11 th and 7 th terms are 5.25 and 3.25 respectively
a + 10 d = 5.25 , a + 6 d = 3.25 , 4 d = 2 , d = ½ a + 5 = 5.25 , a = 0.25 = ¼ , s 15 = 15 / 2 ( 2 * ¼ + 14 * ½ ) = 15 / 2 ( 1 / 2 + 14 / 2 ) = 15 / 2 * 15 / 2 = 225 / 4 = 56.25 answer : a
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "10", "15", "20", "25" ] }
[ "A" ]
in a class , 30 students pass in english and 20 students in math , while some students among these pass in both . how many more students do only english as compared to those doing only maths ?
let us consider tht x student are tohse who passed in both english and maths . . . so first we remove x student from both of them therefore , english = 30 - x maths = 20 - x now , number of students more in english = ( 30 - x ) - ( 20 - x ) = 30 - x - 20 + x = 10 answer : a
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "70", "7", "0.7", "0.07" ] }
[ "D" ]
0.0007 ? = 0.01
"explanation : required answer = 0.0007 / 0.01 = 0.07 / 1 = 0.07 . answer : option d"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "84 m", "88 m", "42 m", "137 m" ] }
[ "C" ]
33 cubic centimetres of silver is drawn into a wire 1 mm in diameter . the length of the wire in metres will be :
"sol . let the length of the wire b h . radius = 1 / 2 mm = 1 / 20 cm . then , 22 / 7 * 1 / 20 * 1 / 20 * h = 33 ⇔ = [ 33 * 20 * 20 * 7 / 22 ] = 4200 cm = 42 m . answer c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "0.000196", "0.00196", "19.6", "196" ] }
[ "A" ]
0.014 × 0.014 = ?
"solution sum of decimals places = 6 . 14 × 14 = 196 . = s 0.014 × 0.014 = 0.000196 answer a"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "3 / 4", "1 / 4", "3 / 8", "7 / 8" ] }
[ "D" ]
three unbiased coins are tossed . what is the probability of getting at most two heads ?
explanation : here s = { ttt , tth , tht , htt , thh , hth , hht , hhh } let e = event of getting at most two heads . then e = { ttt , tth , tht , htt , thh , hth , hht } . p ( e ) = n ( e ) / n ( s ) = 7 / 8 . answer is d
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 15,000", "rs . 15,500", "rs . 12,375", "rs . 16,500" ] }
[ "C" ]
the length of a room is 5.5 m and width is 3.75 m . find the cost of paying the floor by slabs at the rate of rs . 600 per sq . metre .
"solution area of the floor = ( 5.5 x 3.75 ) m ² = 20.635 m ² cost of paying = rs . ( 600 x 20.625 ) = rs . 12,375 . answer c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "16", "1.6", "0.16", "0.016" ] }
[ "B" ]
if 204 ÷ 12.75 = 16 , then 2.04 ÷ 1.275 = ?
explanation : ( 204 / 12.75 ) = 16 ⇒ ( 20.4 ) / ( 1.275 ) = 16 ( ∵ divided numerator and denominator by 10 ) ⇒ 2.04 / 1.275 = 1.6 ( ∵ divided lhs and rhs by 10 ) answer : option b
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "82 m", "50 m", "75 m", "80 m" ] }
[ "C" ]
two trains of equal length are running on parallel lines in the same directions at 46 km / hr . and 36 km / hr . the faster trains pass the slower train in 54 seconds . the length of each train is :
"explanation : the relative speed of train is 46 - 36 = 10 km / hr = ( 10 x 5 ) / 18 = 25 / 9 m / s 10 × 518 = 259 m / s in 54 secs the total distance traveled is 54 x 25 / 9 = 150 m . therefore the length of each train is = 150 / 2 = 75 m . answer c"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "22.8 kg", "26.6 kg", "28 kg", "26.5 kg" ] }
[ "B" ]
if 11.25 m of a uniform steel rod weighs 42.75 kg . what will be the weight of 7 m of the same rod ?
"explanation : let the required weight be x kg . then , less length , less weight ( direct proportion ) = > 11.25 : 7 : : 42.75 : x = > 11.25 x x = 7 x 42.75 = > x = ( 7 x 42.75 ) / 11.25 = > x = 26.6 answer : b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "144 m 2", "140 m 2", "108 m 2", "158 m 2" ] }
[ "B" ]
the floor of a rectangular room is 19 m long and 12 m wide . the room is surrounded by a veranda of width 2 m on all its sides . the area of the veranda is :
"area of the outer rectangle = 23 ã — 16 = 368 m 2 area of the inner rectangle = 19 ã — 12 = 228 m 2 required area = ( 368 â € “ 228 ) = 140 m 2 answer b"
MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "761200", "761400", "761800", "210000" ] }
[ "D" ]
simplify : 500 x 500 - 200 x 200
"( 500 ) ^ 2 - ( 200 ) ^ 2 = ( 500 + 200 ) ( 500 - 200 ) = 700 x 300 = 210000 . answer is d ."
MathQA