shulijia/MNLP_M3_mathqa · Datasets at Hugging Face

choices dict	answerKey listlengths 1 1	question stringlengths 8 751	rationale stringlengths 8 1.53k	dataset stringclasses 1 value
{ "label": [ "A", "B", "C", "D" ], "text": [ "5", "8", "6", "4" ] }	[ "A" ]	‘ a ’ completes a work in 12 days . ‘ b ’ completes the same work in 15 days . ‘ a ’ started working alone and after 3 days b joined him . how many days will they now take together to complete the remaining work ?	work done by ‘ a ’ in 3 days = 1 ⁄ 12 × 3 = 1 ⁄ 4 ∴ remaining work = 1 - 1 ⁄ 4 = 3 ⁄ 4 work done by a and b together = 12 × 15 / 27 = 20 / 3 ∴ remaining work done by a and b together in = 3 ⁄ 4 × 20 ⁄ 3 = 5 days answer a	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "22", "24", "26", "28" ] }	[ "A" ]	in a stream running at 2 kmph , a motor boat goes 6 km upstream and back again to the starting point in 33 minutes . find the speed of the motorboat in still water .	let the speed of the motarboat in still water be x kmph . then , 6 / x + 2 + 6 / x - 2 = 33 / 60 11 x 2 - 240 x - 44 = 0 11 x 2 - 242 x + 2 x - 44 = 0 ( x - 22 ) ( 11 x + 2 ) = 0 x = 22 . answer is a	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs 1.10", "rs 11", "rs 16.50", "rs 27.50" ] }	[ "D" ]	ajay bought 15 kg of dal at the rate of rs 14.50 per kg and 10 kg at the rate of rs 13 per kg . he mixed the two and sold the mixture at the rate of rs 15 per kg . what was his total gain in this transaction ?	explanation : cost price of 25 kg = rs . ( 15 x 14.50 + 10 x 13 ) = rs . 347.50 . sell price of 25 kg = rs . ( 25 x 15 ) = rs . 375 . profit = rs . ( 375 — 347.50 ) = rs . 27.50 . answer : d	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "5000000", "4000000", "1000000", "6000000" ] }	[ "C" ]	p and q invested in a shop . the profits were divided in the ratio of 2 : 4 respectively . if p invested rs . 500000 , the amount invested by q is :	"suppose q invested rs . y . then , 500000 / y = 2 / 4 or y = [ 500000 x 4 / 2 ] = 1000000 . answer c"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "168 °", "134 °", "156 °", "224 °" ] }	[ "B" ]	the ratio of the adjacent angles of a parallelogram is 5 : 13 . also , the ratio of the angles of quadrilateral is 5 : 6 : 7 : 12 . what is the sum of the smaller angle of the parallelogram and the second largest angle of the quadrilateral ?	"the measures of the adjacent angles of a parallelogram add up to be 180 ° given so , 5 x + 13 x = 180 ° or , 18 x = 180 ° or , x = 10 ° hence the angles of the parallelogram are 50 ° and 130 ° further it is given we know sum of all the four angles of a quadrilateral is 360 ° so , 5 y + 6 y + 7 y + 12 y = 360 ° or , 5 y + 6 y + 7 y + 12 y = 360 ° or , 30 y = 360 ° or , y = 12 ° hence the angles of the quadrilateral are 60 ° , 72 , 84 ° and 144 ° will be 50 ° + 84 ° = 134 ° answer : b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "35.2", "36.1", "36.2", "36.4" ] }	[ "D" ]	the mean of 50 observations was 36 . it was found later that an observation 43 was wrongly taken as 23 . the corrected new mean is	"solution correct sum = ( 36 x 50 + 43 - 23 ) = 1820 . â ˆ ´ correct mean = 1820 / 50 = 36.4 . answer d"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "148", "152", "156", "144" ] }	[ "C" ]	the ratio of ducks and frogs in a pond is 37 : 39 respectively . the average number of ducks and frogs in the pond is 152 . what is the number of frogs in the pond ?	"solution : ratio of ducks and frogs in pond , = 37 : 39 . average of ducks and frogs in pond , = 152 . so , total number of ducks and frogs in the pond , = 2 * 152 = 304 . therefore , number of frogs , = ( 304 * 39 ) / 76 = 156 . answer : option c"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "0.6", "0.5", "0.35", "0.3" ] }	[ "D" ]	( 0.8 ) ( power 3 ) - ( 0.5 ) ( power 3 ) / ( 0.8 ) ( power 2 ) + 0.40 + ( 0.5 ) ( power 2 ) is :	"given expression = ( 0.8 ) ( power 3 ) - ( 0.5 ) ( power 3 ) / ( 0.8 ) ( power 2 ) + ( 0.8 x 0.5 ) + ( 0.5 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.8 - 0.5 ) = 0.30 answer is d ."	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "128 m', '", "64 m', '", "64 m ^ 2', '", "128 m ^ 2', '" ] }	[ "B" ]	the area of a parallelogram is 128 sq m . then the area of a triangle formed by its diagonal is - - - - - - - - - - ?	b * h / 2 = 128 / 2 = 64 m . ans : ( b )	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "100", "141.7", "150", "180" ] }	[ "B" ]	one side of a rectangular field is 13 m and one of its diagonal is 17 m . find the area of the field .	"solution other side = √ ( 17 ) 2 - ( 13 ) 2 = √ 289 - 169 = √ 120 = 10.9 m . ∴ area = ( 13 x 10.9 ) m 2 = 141.7 m 2 . answer b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "$ 10800", "$ 18000", "$ 160000", "$ 1800" ] }	[ "C" ]	a man spend 1 / 5 of his salary on food , 1 / 10 of his salary on house rent and 3 / 5 salary on clothes . he still has $ 16000 left with him . find salary . .	"[ 1 / ( x 1 / y 1 + x 2 / y 2 + x 3 / y 3 ) ] * total amount = balance amount [ 1 - ( 1 / 5 + 1 / 10 + 3 / 5 ) } * total salary = $ 16000 , = [ 1 - 9 / 10 ] * total salary = $ 16000 , total salary = $ 16000 * 10 = $ 160000 , correct answer ( c )"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "50", "52", "51", "53" ] }	[ "B" ]	a school has 4 section of chemistry in class x having 60 , 35 , 45 and 42 students . the mean marks obtained in chemistry test are 50 , 60 , 55 and 45 respectively for the 4 sections . determine the overall average of marks per student .	"required average marks = 60 ã — 50 + 35 ã — 60 + 45 ã — 55 + 42 ã — 45 / 60 + 35 + 45 + 42 = 3000 + 2100 + 2475 + 1890 / 182 = 9465 â „ 182 = 52 answer b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "91", "28", "56", "89" ] }	[ "B" ]	the maximum number of student amoung them 1204 pens and 840 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is :	"solution required number of student = h . c . f of 1204 and 840 = 28 . answer b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "57.75 %", "54.54 %", "63 %", "70 %" ] }	[ "A" ]	a cricketer scored 142 runs which included 12 boundaries and 2 sixes . what percent of his total score did he make by running between the wickets .	"explanation : number of runs made by running = 142 - ( 12 x 4 + 2 x 6 ) = 142 - ( 60 ) = 82 now , we need to calculate 82 is what percent of 142 . = > 82 / 142 * 100 = 57.75 % answer : a"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "25 %", "20 %", "24.2 %", "33.33 %" ] }	[ "C" ]	if the price of petrol increases by 32 , 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 + 32 ) i . e 132 . now , his consumption should be reduced to : - = ( 132 − 100 ) / 132 ∗ 100 . hence , the consumption should be reduced to 24.2 % . answer : c	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 1325", "rs . 1300", "rs . 1350", "rs . 1500" ] }	[ "A" ]	the present worth of rs . 1404 due in two equal half - yearly instalments at 8 % per annum simple interest is :	solution required sum = p . w . of rs . 702 due 6 month hence + p . w . of rs . 702 due 1 year hence = rs . [ ( 100 x 702 / 100 + 8 x 1 / 2 ) + ( 100 x 702 / 100 + ( 8 x 1 ) ) ] = rs . ( 675 + 650 ) = rs . 1325 . answer a	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "18 %", "20 %", "25 %", "23 %" ] }	[ "C" ]	two kinds of vodka are mixed in the ratio 1 : 2 and 2 : 1 and they are sold fetching the profit 10 % and 25 % respectively . if the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4 / 3 and 5 / 3 times respectively , then the mixture will fetch the profit of	answer : c .	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "46', '", "81', '", "126', '", "252', '" ] }	[ "C" ]	a rectangular parking space is marked out by painting three of its sides . if the length of the unpainted side is 9 feet , and the sum of the lengths of the painted sides is 37 feet , then what is the area of the parking space in square feet ?	solution clearly , we have : l = 9 and l + 2 b = 37 or b = 14 . ∴ area = ( l × b ) = ( 9 × 14 ) sq . ft . = 126 sq . ft . answer c	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 5870", "rs . 5991", "rs . 6470", "rs . 6850" ] }	[ "C" ]	a grocer has a sale of rs . 5420 , rs . 5660 , rs . 6200 , rs . 6350 and rs . 6500 for 5 consecutive months . find the sale he should have in the sixth month , so that he gets an average sale of rs . 6100 ?	"explanation : total sale for 5 months = rs . ( 5420 + 5660 + 6200 + 6350 + 6500 ) = rs . 30,130 therefore , required sale = rs . [ ( 6100 * 6 ) – 30,130 ] = rs . ( 36600 – 30,130 ) = rs . 6470 answer c"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "38400", "24000", "24936", "25640" ] }	[ "A" ]	60 % of the population of a village is 23040 . the total population of the village is ?	answer ∵ 60 % of p = 23040 ∴ p = ( 23040 x 100 ) / 60 = 38400 correct option : a	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "2 / 3", "3 / 4", "1 / 2", "2 / 8" ] }	[ "A" ]	378 × ? = 252	"explanation : 378 × ? = 252 ? = 252 / 378 = 126 / 189 = 14 / 21 = 2 / 3 answer is a"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "20,20", "20,10", "25,15", "30,10" ] }	[ "D" ]	the ages of two persons differ by 20 years . if 5 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively	explanation : let their ages be x and ( x + 20 ) years . 5 ( x - 5 ) = ( x + 20 - 5 ) or 4 x = 40 or x = 10 . their present ages are 30 years and 10 years option d	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "24', '", "28', '", "36', '", "48', '" ] }	[ "A" ]	a rectangular of certain dimensions is chopped off from one corner of a larger rectangle as shown . ab = 8 cm and bc = 4 cm . the perimeter of the figure abcpqra ( in cm ) is :	solution required perimeter = ( ab + bc + cp + pq + qr + ra ) = ab + bc + ( cp + qr ) + ( pq + ra ) = ab + bc + ab + bc = 2 ( ab + bc ) = [ 2 ( 8 + 4 ) ] cm = 24 cm . answer a	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "1 : 9', '", "1 : 3', '", "3 : 1', '", "1 : 10', '" ] }	[ "A" ]	in a trapezium abcd , ab is parallel to dc , ab = 3 * dc , and the diagonals of the trapezium intersect at o . the ratio of the area of ∆ ocd to the area of ∆ oab is :	triangles are similar , so ratio will be 1 : 9 a	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "30 , 19", "21 , 37", "15 , 34", "15 , 27" ] }	[ "D" ]	the ratio of two numbers is 5 : 9 . if each number is decreased by 5 , the ratio becomes 5 : 11 . find the numbers .	"explanation : let the two numbers be 5 x and 9 x . ( 5 x - 5 ) / ( 9 x - 5 ) = 5 : 11 ( 5 x - 5 ) * 11 = ( 9 x - 5 ) * 5 55 x – 55 = 45 x – 25 10 x = 30 x = 3 therefore , the numbers are 15 and 27 . answer : d"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "171.4", "171.5", "171.6", "171.7" ] }	[ "A" ]	tough and tricky questions : percents . over the course of a year , a certain microbrewery increased its beer output by 20 percent . at the same time , it decreased its total working hours by 30 percent . by what percent did this factory increase its output per hour ?	lets assume the initial production was 100 litres of beer for 100 hr . with the 20 % increase the total amount of beer production will be 120 litres and with 30 % decrease in total hours will be reduced to 70 hr . 100 hr - - - - > 100 lts 1 hr - - - - - > 1 lts 70 hr - - - - - > 120 lts 1 hr - - - - - > 1.714 lts total increase in production for 1 hr = 171.4 % answer a	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "37", "25", "48", "50" ] }	[ "A" ]	the average of 55 results is 28 and the average of other 28 results is 55 . what is the average of all the results ?	"answer sum of 83 result = sum of 55 result + sum of 28 result . = 30 x 20 + 20 x 30 = 3080 correct option : a"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "80 m 2", "84 m 2", "88 m 2", "86 m 2" ] }	[ "C" ]	the diameter of a garden roller is 1.4 m and it is 4 m long . how much area will it cover in 5 revolutions ? ( use ï € = 22 â „ 7 )	"required area covered in 5 revolutions = 5 ã — 2 ï € rh = 5 ã — 2 ã — 22 â „ 7 ã — 0.7 ã — 4 = 88 m 2 answer c"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "0.3408", "3.408", "34.08", "340.8" ] }	[ "D" ]	if 213 × 16 = 3408 , then 16 × 21.3 is equal to :	"solution 16 × 21.3 = ( 16 x 213 / 10 ) = ( 16 x 213 / 10 ) = 3408 / 10 = 340.8 . answer d"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "8", "13", "17", "15" ] }	[ "A" ]	a fraction in reduced form is such that when it is squared and then its numerator is reduced by 33 ( 1 / 3 ) % and denominator is reduced to 20 % , its result is twice the original fraction . the sum of numerator and denominator is :	solution : let the fraction be x / y . when fraction is squared its numerator is reduced by 33 ( 1 / 3 ) and denominator is reduced by 20 % . according to question , ( x / y ) 2 * 33 ( 1 / 3 ) % / 20 % = 2 ( x / y ) . or , ( x / y ) 2 * ( 2 / 3 ) / ( 1 / 5 ) = 2 ( x / y ) . or , x / y = 3 / 5 . sum of numerator and denominator is , ( x + y ) = 3 + 5 = 8 . answer : option a	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "15.92 %", "16.92 %", "17.92 %", "18.92 %" ] }	[ "B" ]	a shopkeeper fixes the marked price of an item 30 % above its cost price . the percentage of discount allowed to gain 8 % is	"explanation : let the cost price = rs 100 then , marked price = rs 130 required gain = 8 % , so selling price = rs 108 discount = 130 - 108 = 22 discount % = ( 22 / 130 ) * 100 = 16.92 % option b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "25", "30", "35", "30" ] }	[ "D" ]	a contractor undertakes to built a walls in 50 days . he employs 20 peoples for the same . however after 25 days he finds that only 40 % of the work is complete . how many more man need to be employed to complete the work in time ?	20 men complete 0.4 work in 25 days . applying the work rule , m 1 × d 1 × w 2 = m 2 × d 2 × w 1 we have , 20 × 25 × 0.6 = m 2 × 25 × 0.4 or m 2 = 20 × 25 × 0.6 / 25 × 0.4 = 30 men answerd	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "45", "30", "24", "40" ] }	[ "A" ]	the ratio of the capacity to do work of a and b is 3 : 2 . if they together can complete a work in 18 days , then how long does a take to complete the work alone ?	llet a and b take 3 x and 2 x days to complete the work 1 / 3 x + 1 / 2 x = 1 / 18 ⇒ x = 15 so a will take 45 days . answer : a	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "40", "77", "48", "52" ] }	[ "B" ]	in a can , there is a mixture of milk and water in the ratio 4 : 5 . if it is filled with an additional 14 litres of milk the can would be full and ratio of milk and water would become 6 : 5 . find the capacity of the can ?	"let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = 4 / 9 ( t - 14 ) after adding milk , quantity of milk in the mixture = 6 / 11 t . 6 t / 11 - 14 = 4 / 9 ( t - 14 ) 10 t = 1386 - 616 = > t = 77 . answer : b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs 8.82", "rs 9.82", "rs 10.82", "rs 14.82" ] }	[ "D" ]	a fruit seller sells mangoes at the rate of rs . 12 per kg and thereby loses 15 % . at what price per kg , he should have sold them to make a profit of 5 %	"explanation : 85 : 12 = 105 : x x = ( 12 × 105 / 85 ) = rs 14.82 option d"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 40", "rs . 50", "rs . 60", "rs . 70" ] }	[ "C" ]	find the simple interest on rs . 5000 at 6 % per annum for the period from 5 th feb to 19 th april , 2015 .	"explanation : given : 1 ) principal = rs . 5000 2 ) rate of interest = 6 % 3 ) time = 5 th feb to 19 th april , 2015 first find the time period 5 th feb to 19 th april , 2015 feb = 28 – 5 = 23 days march = 31 days april = 19 days total days = 23 + 31 + 19 = 73 days convert days into years , by dividing it by 365 time = 73 / 365 = 1 / 5 simple interest = ( p × r × t ) / 100 = [ 5000 × 6 × ( 1 / 5 ) ] / 100 = rs . 60 simple interest = rs . 60 answer is c"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "105", "30", "35", "20" ] }	[ "A" ]	a contractor undertakes to built a walls in 50 days . he employs 70 peoples for the same . however after 25 days he finds that only 40 % of the work is complete . how many more man need to be employed to complete the work in time ?	"70 men complete 0.4 work in 25 days . applying the work rule , m 1 × d 1 × w 2 = m 2 × d 2 × w 1 we have , 70 × 25 × 0.6 = m 2 × 25 × 0.4 or m 2 = 70 × 25 × 0.6 / 25 × 0.4 = 105 men answera"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "1628.4", "110", "1492", "1496" ] }	[ "B" ]	10 people went to a hotel for combine dinner party 8 of them spent rs . 10 each on their dinner and rest spent 4 more than the average expenditure of all the 10 . what was the total money spent by them .	"solution : let average expenditure of 10 people be x . then , 10 x = 8 * 10 + 2 * ( x + 4 ) ; or , 10 x = 10 * 10 + 2 x + 8 ; or , x = 11 ; so , total money spent = 11 * 10 = rs . 110 . answer : option b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "40.3", "40.4", "40.6", "40.8" ] }	[ "A" ]	the average of 10 numbers is 40.2 . later it is found that two numbers have been wrongly copied . the first is 17 greater than the actual number and the second number added is 13 instead of 31 . find the correct average .	"sum of 10 numbers = 402 corrected sum of 10 numbers = 402 – 13 + 31 – 17 = 403 hence , new average = 403 ⁄ 10 = 40.3 answer a"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 15000", "rs . 15550", "rs . 24750", "rs . 16500" ] }	[ "C" ]	the length of a room is 5.5 m and width is 3.75 m . find the cost of paving the floor by slabs at the rate of rs . 1200 per sq . metre .	"solution area of the floor = ( 5.5 × 3.75 ) m 2 = 20.625 m 2 ∴ cost of paving = rs . ( 1200 × 20.625 ) = 24750 . answer c"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "40", "60", "45", "39" ] }	[ "B" ]	a worker makes a toy in every 2 h . if he works for 120 h , then how many toys will he make ?	"no . of toys = 120 / 2 = 60 answer : b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 15,000", "rs . 15,500", "rs . 15,600", "rs . 20,625" ] }	[ "D" ]	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 . 1000 per sq . metre .	"solution area of the floor = ( 5.5 x 3.75 ) m ² = 20.635 m ² cost of paying = rs . ( 1000 x 20.625 ) = rs . 20625 . answer d"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "16', '", "24', '", "32', '", "48', '" ] }	[ "A" ]	the length and breadth of the floor of the room are 20 feet and 10 feet respectively . square tiles of 2 feet length of different colours are to be laid on the floor . black tiles are laid in the first row on all sides . if white tiles are laid in the one - third of the remaining and blue tiles in the rest , how many blue tiles will be there ?	area left after laying black tiles = [ ( 20 – 4 ) × ( 10 – 4 ) ] sq . ft . = 96 sq . ft . area under white tiles = ( 1 ⁄ 3 × 96 ) sq . ft = 32 sq . ft . area under blue tiles = ( 96 – 32 ) sq . ft = 64 sq . ft . number of blue tiles = 64 / 2 × 2 = 16 . answer a	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs 8.5", "rs 9.5", "rs 10.5", "rs 11.5" ] }	[ "C" ]	a fruit seller sells mangoes at the rate of rs . 8 per kg and thereby loses 20 % . at what price per kg , he should have sold them to make a profit of 5 %	explanation : 80 : 8 = 105 : x x = ( 8 × 105 / 80 ) = rs 10.5 option c	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "12,4", "6,3", "9,3", "6,6" ] }	[ "A" ]	a man can row downstream at the rate of 16 km / hr and upstream at 8 km / hr . find man ' s rate in still water and the rate of current ?	"explanation : rate of still water = 1 / 2 ( 16 + 8 ) = 12 km / hr rate of current = 1 / 2 ( 16 - 8 ) = 4 km / hr answer : option a"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "62.2", "2000", "1700", "1729" ] }	[ "D" ]	1852 - 1230 ÷ 10.00 = ?	"answer given expression = 1852 - 1230 ÷ 10.00 = 1852 - 123 = 1729 correct option : d"	MathQA
{ "label": [ "A", "C", "D", "E" ], "text": [ "124 %", "96 %", "80 %", "64 %" ] }	[ "D" ]	mary ' s income is 60 % more than tim ' s income and tim ' s income is 50 % less than juan ' s income . what % of juan ' s income is mary ' s income .	"even i got 96 % j = 100 t = 100 * 0.5 = 50 m = 50 * 1.6 = 80 if mary ' s income is x percent of j m = j * x / 100 x = m * 100 / j = 80 * 100 / 100 = 80 ans : d"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "1008", "1015", "1022", "1032" ] }	[ "B" ]	the smallest number which when diminished by 7 , is divisible 12 , 16 , 18 , 21 and 28 is :	"explanation required number = ( l . c . m . of 12,16 , 18 , 21 , 28 ) + 7 = 1008 + 7 = 1015 answer b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 1200", "rs . 1300", "rs . 1500", "rs . 4000" ] }	[ "D" ]	a man invests rs . 8000 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 × 5 × 1 ) / 100 = rs . 400 let his additional investment at 8 % = x interest on rs . x at 8 % per annum = ( x × 8 × 1 ) / 100 = 2 x / 25 . to earn 6 % per annum for the total , interest = ( 8000 + x ) × 6 × 1 / 100 . = > 400 + 2 x / 25 = ( 8000 + x ) × 6 × 1 / 100 . = > 40000 + 8 x = ( 8000 + x ) × 6 . = > 40000 + 8 x = 48000 + 6 x . = > 2 x = 8000 . = > x = 4000 . answer : d	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "6693.75", "8032.5", "4462.5", "8900" ] }	[ "A" ]	a sum fetched total simple interest of 4016.25 at the rate of 12 p . c . p . a . in 5 years . what is the sum ?	"let the sums be p . now , 60 % of p = 4016.25 or , p = 6693.75 answer a"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "5.3 %", "7.6 %", "10.6 %", "12 %" ] }	[ "B" ]	a man buys 50 pens at marked price of 46 pens from a whole seller . if he sells these pens giving a discount of 1 % , what is the profit percent ?	"explanation : let marked price be re . 1 each c . p . of 50 pens = rs . 46 s . p . of 50 pens = 99 % of rs . 50 = rs . 49.50 profit % = ( profit / c . p . ) x 100 profit % = ( 3.50 / 46 ) x 100 = 7.6 % answer b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "0.35 %", "0.4 %", "0.045 %", "0.6 %" ] }	[ "C" ]	in expressing a length of 61.472 km as nearly as possible with the 3 significant digits , find the percentage error ?	explanation : error = ( 61.5 - 61.472 ) = 0.028 required percentage = 0.028 / 61.472100 = 0.045 answer : c	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "3 .", "2 .", "1 / 5", "1 / 3 ." ] }	[ "C" ]	two brothers took the gmat exam , the higher score is x and the lower one is y . if the difference between the two scores 1 / 4 , what is the value of y / x ?	"answer is c : 1 / 5 x - y = ( x + y ) / 4 solving for y / x = 1 / 5"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "145', '", "135', '", "75', '", "140', '" ] }	[ "B" ]	if the length of rectangle is three times of its breadth . if the area of rectangle is 6075 sq . m , then calculate the length ?	let breadth = x , length = 3 x . area of rectangle = length * breadth = 3 x ^ 2 = 6075 x ^ 2 = 2025 , x = √ 2025 = 45 m length = 135 m answer b	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "40 m 2", "44 m 2", "52.8 m 2", "36 m 2" ] }	[ "C" ]	the diameter of a garden roller is 1.4 m and it is 2 m long . how much area will it cover in 6 revolutions ? ( use ï € = 22 â „ 7 )	"required area covered in 5 revolutions = 6 ã — 2 ï € rh = 6 ã — 2 ã — 22 â „ 7 ã — 0.7 ã — 2 = 52.8 m 2 answer c"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 7000", "rs . 8000", "rs . 8500", "rs . 9000" ] }	[ "B" ]	the salary of a , b , c , d , e is rs . 8000 , rs . 5000 , rs . 11000 , rs . 7000 , rs . 9000 per month respectively , then the average salary of a , b , c , d , and e per month is	"answer average salary = 8000 + 5000 + 11000 + 7000 + 9000 / 5 = rs . 8000 correct option : b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "$ 21", "$ 27", "$ 31", "$ 43" ] }	[ "D" ]	at an upscale fast - food restaurant , shin can buy 3 burgers , 7 shakes , and one cola for $ 120 . at the same place it would cost $ 158.50 for 4 burgers , 10 shakes , and one cola . how much would it cost for a meal of one burger , one shake , and one cola ?	"let ' s suppose that the price of a burger is bb , of a shake - ss and that of a cola is cc . we can then construct these equations : 3 b + 7 s + c = 120 4 b + 10 s + c = 158.5 subtracting the first equation from the second gives us b + 3 s = 38.5 now if we subtract the new equation two times from first or 3 times from second we will get b + s + c = 43 . in any case , there is no necessity to know each item ' s price , just the sum . answer : d"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 72", "rs . 65", "rs . 54", "rs . 50" ] }	[ "B" ]	the banker ' s gain on a bill due due 1 year hence at 12 % per annum is rs . 7.8 . the true discount is	"solution t . d = [ b . g x 100 / r x t ] = rs . ( 7.8 x 100 / 12 x 1 ) = rs . 65 . answer b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 55.50", "rs . 67.50", "rs . 86.50", "rs . 101.50" ] }	[ "D" ]	the sides of a rectangular field are in the ratio 3 : 4 . if the area of the field is 10092 sq . m , the cost of fencing the field @ 25 paise per metre is	"solution let length = ( 3 x ) metres and breadth = ( 4 x ) metres . then , 3 x × 4 x = 10092 ⇔ 12 x 2 = 10092 ⇔ x 2 = 841 ⇔ x = 29 . so , length = 87 m and breadth = 116 m . perimeter = [ 2 ( 87 + 116 ) ] m = 406 m . ∴ cost of fencing = rs . ( 0.25 × 406 ) = rs . 101.50 . answer d"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 147.50", "rs . 785.50", "rs . 174.50", "rs . 258.50" ] }	[ "C" ]	teas worth rs . 126 per kg and rs . 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs 152 per kg , the price of the third variety per kg will be ?	"explanation : since first and second varieties are mixed in equal proportions . so , their average price = rs . ( 126 + 135 ) / 2 . = > rs . 130.50 . so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . by the rule of alligation , we have : cost of 1 kg cost of 1 kg of 1 st kind of 2 nd kind ( rs . 130.50 ) ( rs . x ) \ / mean price ( rs . 152 ) / \ x â ˆ ’ 152 22.50 = > x â ˆ ’ ( 152 / 22.50 ) = 1 . = > x â ˆ ’ 152 = 22.50 . = > x = 174.50 rs . answer : c"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "11236', '", "11025', '", "14400', '", "12696', '" ] }	[ "A" ]	a lady grows cabbage in her garden that is in the shape of a square . each cabbage takes 1 square foot of area in her garden . this year , she has increased her output by 211 cabbages when compared to last year . the shape of the area used for growing the cabbage has remained a square in both these years . how many cabbages did she produce this year ?	explanatory answer the shape of the area used for growing cabbages has remained a square in both the years . let the side of the square area used for growing cabbages this year be x ft . therefore , the area of the ground used for cultivation this year = x 2 sq . ft . let the side of the square area used for growing cabbages last year be y ft . therefore , the area of the ground used for cultivation last year = y 2 sq . ft . as the number of cabbages grown has increased by 211 , the area would have increased by 211 sq ft because each cabbage takes 1 sq ft space . hence , x 2 - y 2 = 211 ( x + y ) ( x - y ) = 211 . 211 is a prime number and hence it will have only two factors . i . e . , 211 and 1 . therefore , 211 can be expressed as product of 2 numbers in only way = 211 * 1 i . e . , ( x + y ) ( x - y ) = 211 * 1 so , ( x + y ) should be 211 and ( x - y ) should be 1 . solving the two equations we get x = 106 and y = 105 . therefore , number of cabbages produced this year = x 2 = 1062 = 11236 . alternative approach : use answer choices the area in both the years are squares of two numbers . that rules out choice d . 12696 is not the square of any number . check choice a : if this year ' s produce is 11236 , last year ' s produce would have been 11236 - 211 = 11025 11025 is the square of 105 . so , 11236 is the answer . choice a is the correct answer .	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "130 km", "120 km", "100 km", "280 km" ] }	[ "D" ]	a bus covered a distance of 250 km , partly at an average speed of 40 kmph and partly at 60 kmph . if the total time taken is 6.5 hours , then the distance covered at 40 kmph is	"let the partial distance covered at 40 kmph be x let the another partial distance covered at 60 kmph be ( 250 - x ) thus , x / 40 - ( 250 - x ) / 60 = 6.5 or , x / 40 + ( 250 - x ) / 60 = 6.5 or , ( 3 x + 500 - 2 x ) / / 120 = 6.5 or 500 + x = 780 x = 280 answer : d"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "10.22 %", "20.22 %", "21.22 %", "40 %" ] }	[ "D" ]	a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 12 % while buying and by 20 % while selling . what is his percentage profit ?	"the owner buys 100 kg but actually gets 112 kg ; the owner sells 100 kg but actually gives 80 kg ; profit : ( 112 - 80 ) / 80 * 100 = ~ 40 % answer : d ."	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "48", "68", "54", "60" ] }	[ "C" ]	70 % of 300 + 25 % of 400 - ? = 256	"explanation : solution : let 70 % of 300 + 25 % of 400 - x = 256 . then , x = ( 70 / 100 * 300 ) + ( 25 / 100 * 400 ) - 256 = 210 + 100 - 256 = 54 . answer : c"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "95 %", "93 %", "90 %", "80 %" ] }	[ "B" ]	two numbers are respectively 40 % and 30 % more than a third number . the second number expressed in terms of percentage of the first is ?	"here , x = 40 and y = 30 ; therefore second number = [ [ ( 100 + y ) / ( 100 + x ) ] x 100 ] % of first number = [ [ ( 100 + 30 ) / ( 100 + 40 ) ] x 100 ] % of first number = 92.8 % of the first answer : b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "13 , 3", "18 , 4", "15 , 3", "14 , 4" ] }	[ "B" ]	a man can row downstream at 22 kmph and upstream at 14 kmph . find the speed of the man in still water and the speed of stream respectively ?	"explanation : let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 22 - - - ( 1 ) and x - y = 14 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 36 = > x = 18 , y = 4 . answer : option b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "45 ( 4 / 11 ) %", "50 %", "45 ( 5 / 11 ) %", "44 ( 5 / 11 ) %" ] }	[ "B" ]	a batsman scored 120 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 = 120 total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60 total runs scored by running between the wickets = 120 - 60 = 60 required % = ( 60 / 120 ) × 100 = 50 % answer : option b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "25 %", "50 %", "75 %", "95 %" ] }	[ "C" ]	if the radius of a circle is decreased by 50 % , find the percentage decrease in its area .	"let original radius = r . new radius = ( 50 / 100 ) r = ( r / 2 ) original area = 22 / 7 ( r ) 2 = and new area =  ( ( r / 2 ) ) 2 = (  ( r ) 2 ) / 4 decrease in area = ( ( 3 ( 22 / 7 ) ( r ) 2 ) / 4 x ( 1 / ( 22 / 7 ) ( r ) 2 ) x 100 ) % = 75 % answer is c ."	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "0.1602", "0.001602", "1.6021", "0.01602" ] }	[ "D" ]	16.02 × 0.001 = ?	"16.02 × 0.001 = ? or , ? = 0.01602 answer d"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 5725", "rs . 5275", "rs . 6275", "rs . 6500" ] }	[ "D" ]	the owner of a furniture shop charges his customer 30 % more than the cost price . if a customer paid rs . 8450 for a computer table , then what was the cost price of the computer table ?	"cp = sp * ( 100 / ( 100 + profit % ) ) = 8450 ( 100 / 130 ) = rs . 6500 . answer : d"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "10", "5", "9", "15" ] }	[ "C" ]	the sum of the squares of three numbers is 41 , while the sum of their products taken two at a time is 20 . their sum is :	"x ^ + y ^ 2 + z ^ 2 = 138 xy + yz + zx = 131 as we know . . ( x + y + z ) ^ 2 = x ^ 2 + y ^ 2 + z ^ 2 + 2 ( xy + yz + zx ) so ( x + y + z ) ^ 2 = 41 + ( 2 * 20 ) ( x + y + z ) ^ 2 = 81 so x + y + z = 9 answer : c"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 169.50", "rs . 1700", "rs . 195.50", "rs . 180" ] }	[ "C" ]	tea worth rs . 126 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs . 173 per kg , the price of the third variety per kg will be	"solution since first second varieties are mixed in equal proportions , so their average price = rs . ( 126 + 135 / 2 ) = rs . 130.50 so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . x - 173 / 22.50 = 1 = â € º x - 173 = 22.50 = â € º x = 195.50 . hence , price of the third variety = rs . 195.50 per kg . answer c"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "13000", "7000", "10000", "5000" ] }	[ "B" ]	a textile manufacturing firm employees 50 looms . it makes fabrics for a branded company . the aggregate sales value of the output of the 50 looms is rs 5 , 00,000 and the monthly manufacturing expenses is rs 1 , 50,000 . 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 − ( 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 × ( 49 / 50 ) − 150000 × ( 49 / 50 ) − 75000 . = > rs 2 , 68,000 . decrease in profit = > 2 , 75,000 − 2 , 68,000 = > rs . 7,000 . answer : b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "356000", "356500", "357000", "400000" ] }	[ "C" ]	in an election , candidate a got 75 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favour of candidate	"total number of invalid votes = 15 % of 560000 = 15 / 100 × 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 – 84000 = 476000 percentage of votes polled in favour of candidate a = 75 % therefore , the number of valid votes polled in favour of candidate a = 75 % of 476000 = 75 / 100 × 476000 = 35700000 / 100 = 357000 c )"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "4000", "3050", "4400", "4500" ] }	[ "A" ]	if 15 % of 30 % of 50 % of a number is 90 , then what is the number ?	"let the number be a given , 15 / 100 * 30 / 100 * 50 / 100 * a = 90 = > 3 / 20 * 3 / 10 * 1 / 2 * a = 90 = > a = 10 * 20 * 10 * 2 = 4000 . answer : 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 . answer b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "14 / 15", "15 / 14", "4 / 5", "42 / 55" ] }	[ "D" ]	jack and jill are marathon runners . jack can finish a marathon ( 42 km ) in 5.5 hours and jill can run a marathon in 4.2 hours . what is the ratio of their average running speed ? ( jack : jill )	"average speed of jack = distance / time = 42 / ( 11 / 2 ) = 84 / 11 average speed of jill = 42 / ( 4.2 ) = 10 ratio of average speed of jack to jill = ( 84 / 11 ) / 10 = 84 / 110 = 42 / 55 answer d"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "40", "50", "200", "120" ] }	[ "C" ]	the length of a rectangulat plot is 20 metres more than its breadth . if the cost of fencing the plot @ rs . 26.50 per metre is rs . 5300 , what is the length of the plot in metres ?	solution let breadth = x metres then , length = ( x + 20 ) metres . perimeter = ( 5300 / 26.50 ) m = 200 m answer c	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "75", "100", "125", "180" ] }	[ "C" ]	the difference between a number and its three fifth is 50 . what is the number ?	"solution let the number be x . then , x - 3 / 5 x = 50 ‹ = › 2 / 5 x = 50 ‹ = › x = ( 50 x 5 / 2 ) ‹ = › 125 . answer c"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "55", "56", "14", "58" ] }	[ "C" ]	a rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing . if the poles of the fence are kept 20 metres apart , how many poles will be needed ?	"solution perimeter of the plot = 2 ( 90 + 50 ) = 280 m . ∴ number of poles = [ 280 / 20 ] = 14 m answer c"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ ". 0028", ". 0027", ". 0026", ". 0025" ] }	[ "B" ]	what decimal of an hour is a second	explanation : 1 / ( 60 * 60 ) = 1 / 3600 = . 0027 option b	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs 20123.20", "rs 20246.4", "rs 20123.40", "rs 20123.50" ] }	[ "B" ]	what will be the compound interest on rs . 50000 after 3 years at the rate of 12 % per annum	"explanation : ( 50000 × ( 1 + 12 / 100 ) 3 ) = > 50000 × 28 / 25 × 28 / 25 × 28 / 25 = > 70246.4 so compound interest will be 70246.4 - 50000 = rs 20246.4 option b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "243", "253", "312", "432" ] }	[ "B" ]	when a number is divided by 13 , the remainder is 6 . when the same number is divided by 7 , then remainder is 1 . what is the number ?	"explanation : take 243 243 ÷ 7 = 34 , remainder = 5 hence this is not the answer take 312 312 ÷ 7 = 44 , remainder = 4 hence this is not the answer take 253 253 ÷ 7 = 36 , remainder = 1 . 253 ÷ 13 = 19 , remainder = 6 this satisfies both the conditions given in the question . hence it is the answer . answer : b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 15250", "rs . 13375", "rs . 16750", "rs . 18000" ] }	[ "D" ]	a , b and c enter into a partnership . a invests 3 times as much as b invests and 2 / 3 of what c invests . at the end of the year , the profit earned is rs . 66000 . what is the share of c ?	"explanation : let the investment of c be rs . x . the inverstment of b = rs . ( 2 x / 3 ) the inverstment of a = rs . ( 3 × ( 2 / 3 ) x ) = rs . ( 2 x ) ratio of capitals of a , b and c = 2 x : 2 x / 3 : x = 6 : 2 : 3 c ' s share = rs . [ ( 3 / 11 ) × 66000 ] = rs . 18000 answer : option d"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "50.12", "52.12", "51.12", "53.12" ] }	[ "B" ]	a school has 4 section of chemistry in class x having 50 , 35 , 45 and 42 students . the mean marks obtained in chemistry test are 50 , 60 , 55 and 45 respectively for the 4 sections . determine the overall average of marks per student .	"required average marks = 50 ã — 50 + 35 ã — 60 + 45 ã — 55 + 42 ã — 45 / 50 + 35 + 45 + 42 = 2500 + 2100 + 2475 + 1890 / 172 = 8965 â „ 172 = 52.12 answer b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "10 %", "9 %", "11.11 %", "42.85 %" ] }	[ "D" ]	a shopkeeper forced to sell at cost price , uses a 700 grams weight for a kilogram . what is his gain percent ?	"shopkeeper sells 700 g instead of 1000 g . so , his gain = 1000 - 700 = 300 g . thus , % gain = ( 300 * 100 ) / 700 = 42.85 % . answer : option d"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "1", "4", "81", "log 80 243" ] }	[ "B" ]	what is log 3 ( 4 ) log 4 ( 5 ) . . . log 80 ( 81 ) ?	recall the change of base formula logb ( a ) = ln ( a ) ln ( b ) : ( alternatively , we can substitute logc for ln on the right , as long as the base is the same on the top and on the bottom . ) using this , we can rewrite the entire expression using natural logarithms , as ln ( 4 ) ln ( 3 ) ln ( 5 ) ln ( 4 ) ln ( 81 ) ln ( 80 ) = ln ( 81 ) ln ( 3 ) = log 3 ( 81 ) = log 3 ( 34 ) = 4 : correct answer b	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 300", "rs . 200", "rs . 240", "rs . 350" ] }	[ "B" ]	if x gets 25 % more than y and y gets 20 % more than z , the share of z out of rs . 740 will be :	"z share = z , y = 1.2 z x = 1.25 × 1.2 z , x + y + z = 740 ( 1.25 × 1.2 + 1.2 + 1 ) z = 74 3.7 z = 740 , z = 200 answer : . b ."	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 660", "rs . 760", "rs . 600", "rs . 960" ] }	[ "C" ]	a man buys an item at rs . 750 and sells it at the loss of 20 percent . then what is the selling price of that item	"explanation : here always remember , when ever x % loss , it means s . p . = ( 100 - x ) % of c . p when ever x % profit , it means s . p . = ( 100 + x ) % of c . p so here will be ( 100 - x ) % of c . p . = 80 % of 750 = 80 / 100 * 750 = 600 option c"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "20,000", "21,200", "1,800", "20,500" ] }	[ "C" ]	a , b and c start a business each investing 2,000 . after 6 months a withdrew 1000 , b withdrew 1000 and c invests 1000 more . at the end of the year , a total profit of 6600 was recorded . find the share of b .	"ratio of the capitals of a , b and c = 2000 ã — 6 + 1000 ã — 6 : 2000 ã — 6 + 1000 ã — 6 : 2000 ã — 6 + 3000 ã — 6 = 18000 : 18000 : 30000 = 18 : 18 : 30 b â € ™ s share = ( 6600 ã — 18 â „ 66 ) = 1800 answer c"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 4000", "rs . 6000", "rs . 7500", "rs . 6600" ] }	[ "B" ]	3 partners a , b , c starts a business . twice a ' s capital is equal to thrice b ' s capital and b ' s capital is 4 times c ' s capital . out of a total profit of rs . 16500 at the end of the year , b ' s share is	solution let c = x . then , b = 4 x and 2 a = 3 x 4 x = 12 x or a = 6 x . ∴ a : b : c = 6 x : 4 x : x = 6 : 4 : 1 . so b ' s capital = rs ( 16500 x 4 / 11 ) = rs . 6000 . answer b	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "5 kg", "15 kg", "25 kg", "50 kg" ] }	[ "D" ]	the price of rice falls by 20 % . how much rice can be bought now with the money that was sufficient to buy 40 kg of rice previously ?	"solution : let rs . 100 be spend on rice initially for 40 kg . as the price falls by 20 % , new price for 40 kg rice , = ( 100 - 20 % of 100 ) = 80 new price of rice = 80 / 40 = rs . 2 per kg . rice can bought now at = 100 / 2 = 50 kg . answer : option d"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "50", "2.5", "25", ". 25" ] }	[ "A" ]	1 / 0.02 is equal to	explanation : 1 / 0.02 = ( 1 * 100 ) / 2 = 100 / 2 = 50 option a	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "7.5 hr", "8 hr", "8.5 hr", "10 hr" ] }	[ "A" ]	two pipes can fill the cistern in 10 hr and 12 hr respectively , while the third empty it in 20 hr . if all pipes are opened simultaneously , then the cistern will be filled in	"solution : work done by all the tanks working together in 1 hour . 1 / 10 + 1 / 12 − 1 / 20 = 2 / 15 hence , tank will be filled in 15 / 2 = 7.5 hour option ( a )"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "22725", "25675", "22655", "27575" ] }	[ "A" ]	2525 * 9	"explanation : 2525 * ( 10 - 1 ) = 25250 - 2525 = 22725 option a"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 150.40", "rs . 44.80", "rs . 140.40", "rs . 50.48" ] }	[ "B" ]	if the sales tax be reduced from 3 ( 1 / 5 ) % to 2 ( 1 / 2 ) % , then what difference does it make to a person who purchases a bag with marked price of rs . 6400 ?	"explanation : required difference = ( 3 ( 1 / 5 ) of rs . 6400 ) - ( 2 ( 1 / 2 ) of rs . 6400 ) = ( 16 / 5 – 5 / 2 ) % of rs . 6400 = ( 7 / 10 ) x ( 1 / 100 ) x 6400 = rs . 44.80 answer b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 5870", "rs . 5991", "rs . 6020", "rs . 7670" ] }	[ "D" ]	a grocer has a sale of rs . 5420 , rs . 5660 , rs . 6200 , rs . 6350 and rs . 6500 for 5 consecutive months . find the sale he should have in the sixth month , so that he gets an average sale of rs . 6300 ?	"explanation : total sale for 5 months = rs . ( 5420 + 5660 + 6200 + 6350 + 6500 ) = rs . 30,130 therefore , required sale = rs . [ ( 6300 * 6 ) – 30,130 ] = rs . ( 37800 – 30,130 ) = rs . 7670 answer d"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "60.8", "18", "30", "52.5" ] }	[ "D" ]	a train moves with a speed of 189 kmph . its speed in metres per second is	"solution speed = 189 kmph = ( 189 x 5 / 18 ) m / sec = 52.5 m / sec . answer d"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "148 min .", "150 min .", "145 min .", "155 min ." ] }	[ "B" ]	a man is walking at a speed of 10 km per hour . after every kilometre , he takes rest for 10 minutes . how much time will be take to cover a distance of 10 kilometres ?	"rest time = number of rest ã — time for each rest = 9 ã — 10 = 90 minutes total time to cover 10 km = ( 10 â „ 10 ã — 60 ) minutes + 90 minutes = 150 minutes answer b"	MathQA
{ "label": [ "A", "B", "C", "D" ], "text": [ "111", "1100", "121", "100" ] }	[ "C" ]	find 10 / 11 = 110 / ?	"answer let 10 / 11 = 110 / n , then , 10 n = 11 x 110 ∴ n = ( 11 x 110 ) / 10 = 121 . option : c"	MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "5", "8", "6", "4" ] }

[ "A" ]

‘ a ’ completes a work in 12 days . ‘ b ’ completes the same work in 15 days . ‘ a ’ started working alone and after 3 days b joined him . how many days will they now take together to complete the remaining work ?

work done by ‘ a ’ in 3 days = 1 ⁄ 12 × 3 = 1 ⁄ 4 ∴ remaining work = 1 - 1 ⁄ 4 = 3 ⁄ 4 work done by a and b together = 12 × 15 / 27 = 20 / 3 ∴ remaining work done by a and b together in = 3 ⁄ 4 × 20 ⁄ 3 = 5 days answer a

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "22", "24", "26", "28" ] }

[ "A" ]

in a stream running at 2 kmph , a motor boat goes 6 km upstream and back again to the starting point in 33 minutes . find the speed of the motorboat in still water .

let the speed of the motarboat in still water be x kmph . then , 6 / x + 2 + 6 / x - 2 = 33 / 60 11 x 2 - 240 x - 44 = 0 11 x 2 - 242 x + 2 x - 44 = 0 ( x - 22 ) ( 11 x + 2 ) = 0 x = 22 . answer is a

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs 1.10", "rs 11", "rs 16.50", "rs 27.50" ] }

[ "D" ]

ajay bought 15 kg of dal at the rate of rs 14.50 per kg and 10 kg at the rate of rs 13 per kg . he mixed the two and sold the mixture at the rate of rs 15 per kg . what was his total gain in this transaction ?

explanation : cost price of 25 kg = rs . ( 15 x 14.50 + 10 x 13 ) = rs . 347.50 . sell price of 25 kg = rs . ( 25 x 15 ) = rs . 375 . profit = rs . ( 375 — 347.50 ) = rs . 27.50 . answer : d

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "5000000", "4000000", "1000000", "6000000" ] }

[ "C" ]

p and q invested in a shop . the profits were divided in the ratio of 2 : 4 respectively . if p invested rs . 500000 , the amount invested by q is :

"suppose q invested rs . y . then , 500000 / y = 2 / 4 or y = [ 500000 x 4 / 2 ] = 1000000 . answer c"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "168 °", "134 °", "156 °", "224 °" ] }

[ "B" ]

the ratio of the adjacent angles of a parallelogram is 5 : 13 . also , the ratio of the angles of quadrilateral is 5 : 6 : 7 : 12 . what is the sum of the smaller angle of the parallelogram and the second largest angle of the quadrilateral ?

"the measures of the adjacent angles of a parallelogram add up to be 180 ° given so , 5 x + 13 x = 180 ° or , 18 x = 180 ° or , x = 10 ° hence the angles of the parallelogram are 50 ° and 130 ° further it is given we know sum of all the four angles of a quadrilateral is 360 ° so , 5 y + 6 y + 7 y + 12 y = 360 ° or , 5 y + 6 y + 7 y + 12 y = 360 ° or , 30 y = 360 ° or , y = 12 ° hence the angles of the quadrilateral are 60 ° , 72 , 84 ° and 144 ° will be 50 ° + 84 ° = 134 ° answer : b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "35.2", "36.1", "36.2", "36.4" ] }

[ "D" ]

the mean of 50 observations was 36 . it was found later that an observation 43 was wrongly taken as 23 . the corrected new mean is

"solution correct sum = ( 36 x 50 + 43 - 23 ) = 1820 . â ˆ ´ correct mean = 1820 / 50 = 36.4 . answer d"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "148", "152", "156", "144" ] }

[ "C" ]

the ratio of ducks and frogs in a pond is 37 : 39 respectively . the average number of ducks and frogs in the pond is 152 . what is the number of frogs in the pond ?

"solution : ratio of ducks and frogs in pond , = 37 : 39 . average of ducks and frogs in pond , = 152 . so , total number of ducks and frogs in the pond , = 2 * 152 = 304 . therefore , number of frogs , = ( 304 * 39 ) / 76 = 156 . answer : option c"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "0.6", "0.5", "0.35", "0.3" ] }

[ "D" ]

( 0.8 ) ( power 3 ) - ( 0.5 ) ( power 3 ) / ( 0.8 ) ( power 2 ) + 0.40 + ( 0.5 ) ( power 2 ) is :

"given expression = ( 0.8 ) ( power 3 ) - ( 0.5 ) ( power 3 ) / ( 0.8 ) ( power 2 ) + ( 0.8 x 0.5 ) + ( 0.5 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.8 - 0.5 ) = 0.30 answer is d ."

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "128 m', '", "64 m', '", "64 m ^ 2', '", "128 m ^ 2', '" ] }

[ "B" ]

the area of a parallelogram is 128 sq m . then the area of a triangle formed by its diagonal is - - - - - - - - - - ?

b * h / 2 = 128 / 2 = 64 m . ans : ( b )

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "100", "141.7", "150", "180" ] }

[ "B" ]

one side of a rectangular field is 13 m and one of its diagonal is 17 m . find the area of the field .

"solution other side = √ ( 17 ) 2 - ( 13 ) 2 = √ 289 - 169 = √ 120 = 10.9 m . ∴ area = ( 13 x 10.9 ) m 2 = 141.7 m 2 . answer b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "$ 10800", "$ 18000", "$ 160000", "$ 1800" ] }

[ "C" ]

a man spend 1 / 5 of his salary on food , 1 / 10 of his salary on house rent and 3 / 5 salary on clothes . he still has $ 16000 left with him . find salary . .

"[ 1 / ( x 1 / y 1 + x 2 / y 2 + x 3 / y 3 ) ] * total amount = balance amount [ 1 - ( 1 / 5 + 1 / 10 + 3 / 5 ) } * total salary = $ 16000 , = [ 1 - 9 / 10 ] * total salary = $ 16000 , total salary = $ 16000 * 10 = $ 160000 , correct answer ( c )"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "50", "52", "51", "53" ] }

[ "B" ]

a school has 4 section of chemistry in class x having 60 , 35 , 45 and 42 students . the mean marks obtained in chemistry test are 50 , 60 , 55 and 45 respectively for the 4 sections . determine the overall average of marks per student .

"required average marks = 60 ã — 50 + 35 ã — 60 + 45 ã — 55 + 42 ã — 45 / 60 + 35 + 45 + 42 = 3000 + 2100 + 2475 + 1890 / 182 = 9465 â „ 182 = 52 answer b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "91", "28", "56", "89" ] }

[ "B" ]

the maximum number of student amoung them 1204 pens and 840 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is :

"solution required number of student = h . c . f of 1204 and 840 = 28 . answer b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "57.75 %", "54.54 %", "63 %", "70 %" ] }

[ "A" ]

a cricketer scored 142 runs which included 12 boundaries and 2 sixes . what percent of his total score did he make by running between the wickets .

"explanation : number of runs made by running = 142 - ( 12 x 4 + 2 x 6 ) = 142 - ( 60 ) = 82 now , we need to calculate 82 is what percent of 142 . = > 82 / 142 * 100 = 57.75 % answer : a"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "25 %", "20 %", "24.2 %", "33.33 %" ] }

[ "C" ]

if the price of petrol increases by 32 , 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 + 32 ) i . e 132 . now , his consumption should be reduced to : - = ( 132 − 100 ) / 132 ∗ 100 . hence , the consumption should be reduced to 24.2 % . answer : c

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 1325", "rs . 1300", "rs . 1350", "rs . 1500" ] }

[ "A" ]

the present worth of rs . 1404 due in two equal half - yearly instalments at 8 % per annum simple interest is :

solution required sum = p . w . of rs . 702 due 6 month hence + p . w . of rs . 702 due 1 year hence = rs . [ ( 100 x 702 / 100 + 8 x 1 / 2 ) + ( 100 x 702 / 100 + ( 8 x 1 ) ) ] = rs . ( 675 + 650 ) = rs . 1325 . answer a

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "18 %", "20 %", "25 %", "23 %" ] }

[ "C" ]

two kinds of vodka are mixed in the ratio 1 : 2 and 2 : 1 and they are sold fetching the profit 10 % and 25 % respectively . if the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4 / 3 and 5 / 3 times respectively , then the mixture will fetch the profit of

answer : c .

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "46', '", "81', '", "126', '", "252', '" ] }

[ "C" ]

a rectangular parking space is marked out by painting three of its sides . if the length of the unpainted side is 9 feet , and the sum of the lengths of the painted sides is 37 feet , then what is the area of the parking space in square feet ?

solution clearly , we have : l = 9 and l + 2 b = 37 or b = 14 . ∴ area = ( l × b ) = ( 9 × 14 ) sq . ft . = 126 sq . ft . answer c

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 5870", "rs . 5991", "rs . 6470", "rs . 6850" ] }

[ "C" ]

a grocer has a sale of rs . 5420 , rs . 5660 , rs . 6200 , rs . 6350 and rs . 6500 for 5 consecutive months . find the sale he should have in the sixth month , so that he gets an average sale of rs . 6100 ?

"explanation : total sale for 5 months = rs . ( 5420 + 5660 + 6200 + 6350 + 6500 ) = rs . 30,130 therefore , required sale = rs . [ ( 6100 * 6 ) – 30,130 ] = rs . ( 36600 – 30,130 ) = rs . 6470 answer c"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "38400", "24000", "24936", "25640" ] }

[ "A" ]

60 % of the population of a village is 23040 . the total population of the village is ?

answer ∵ 60 % of p = 23040 ∴ p = ( 23040 x 100 ) / 60 = 38400 correct option : a

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "2 / 3", "3 / 4", "1 / 2", "2 / 8" ] }

[ "A" ]

378 × ? = 252

"explanation : 378 × ? = 252 ? = 252 / 378 = 126 / 189 = 14 / 21 = 2 / 3 answer is a"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "20,20", "20,10", "25,15", "30,10" ] }

[ "D" ]

the ages of two persons differ by 20 years . if 5 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively

explanation : let their ages be x and ( x + 20 ) years . 5 ( x - 5 ) = ( x + 20 - 5 ) or 4 x = 40 or x = 10 . their present ages are 30 years and 10 years option d

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "24', '", "28', '", "36', '", "48', '" ] }

[ "A" ]

a rectangular of certain dimensions is chopped off from one corner of a larger rectangle as shown . ab = 8 cm and bc = 4 cm . the perimeter of the figure abcpqra ( in cm ) is :

solution required perimeter = ( ab + bc + cp + pq + qr + ra ) = ab + bc + ( cp + qr ) + ( pq + ra ) = ab + bc + ab + bc = 2 ( ab + bc ) = [ 2 ( 8 + 4 ) ] cm = 24 cm . answer a

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "1 : 9', '", "1 : 3', '", "3 : 1', '", "1 : 10', '" ] }

[ "A" ]

in a trapezium abcd , ab is parallel to dc , ab = 3 * dc , and the diagonals of the trapezium intersect at o . the ratio of the area of ∆ ocd to the area of ∆ oab is :

triangles are similar , so ratio will be 1 : 9 a

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "30 , 19", "21 , 37", "15 , 34", "15 , 27" ] }

[ "D" ]

the ratio of two numbers is 5 : 9 . if each number is decreased by 5 , the ratio becomes 5 : 11 . find the numbers .

"explanation : let the two numbers be 5 x and 9 x . ( 5 x - 5 ) / ( 9 x - 5 ) = 5 : 11 ( 5 x - 5 ) * 11 = ( 9 x - 5 ) * 5 55 x – 55 = 45 x – 25 10 x = 30 x = 3 therefore , the numbers are 15 and 27 . answer : d"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "171.4", "171.5", "171.6", "171.7" ] }

[ "A" ]

tough and tricky questions : percents . over the course of a year , a certain microbrewery increased its beer output by 20 percent . at the same time , it decreased its total working hours by 30 percent . by what percent did this factory increase its output per hour ?

lets assume the initial production was 100 litres of beer for 100 hr . with the 20 % increase the total amount of beer production will be 120 litres and with 30 % decrease in total hours will be reduced to 70 hr . 100 hr - - - - > 100 lts 1 hr - - - - - > 1 lts 70 hr - - - - - > 120 lts 1 hr - - - - - > 1.714 lts total increase in production for 1 hr = 171.4 % answer a

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "37", "25", "48", "50" ] }

[ "A" ]

the average of 55 results is 28 and the average of other 28 results is 55 . what is the average of all the results ?

"answer sum of 83 result = sum of 55 result + sum of 28 result . = 30 x 20 + 20 x 30 = 3080 correct option : a"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "80 m 2", "84 m 2", "88 m 2", "86 m 2" ] }

[ "C" ]

the diameter of a garden roller is 1.4 m and it is 4 m long . how much area will it cover in 5 revolutions ? ( use ï € = 22 â „ 7 )

"required area covered in 5 revolutions = 5 ã — 2 ï € rh = 5 ã — 2 ã — 22 â „ 7 ã — 0.7 ã — 4 = 88 m 2 answer c"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "0.3408", "3.408", "34.08", "340.8" ] }

[ "D" ]

if 213 × 16 = 3408 , then 16 × 21.3 is equal to :

"solution 16 × 21.3 = ( 16 x 213 / 10 ) = ( 16 x 213 / 10 ) = 3408 / 10 = 340.8 . answer d"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "8", "13", "17", "15" ] }

[ "A" ]

a fraction in reduced form is such that when it is squared and then its numerator is reduced by 33 ( 1 / 3 ) % and denominator is reduced to 20 % , its result is twice the original fraction . the sum of numerator and denominator is :

solution : let the fraction be x / y . when fraction is squared its numerator is reduced by 33 ( 1 / 3 ) and denominator is reduced by 20 % . according to question , ( x / y ) 2 * 33 ( 1 / 3 ) % / 20 % = 2 ( x / y ) . or , ( x / y ) 2 * ( 2 / 3 ) / ( 1 / 5 ) = 2 ( x / y ) . or , x / y = 3 / 5 . sum of numerator and denominator is , ( x + y ) = 3 + 5 = 8 . answer : option a

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "15.92 %", "16.92 %", "17.92 %", "18.92 %" ] }

[ "B" ]

a shopkeeper fixes the marked price of an item 30 % above its cost price . the percentage of discount allowed to gain 8 % is

"explanation : let the cost price = rs 100 then , marked price = rs 130 required gain = 8 % , so selling price = rs 108 discount = 130 - 108 = 22 discount % = ( 22 / 130 ) * 100 = 16.92 % option b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "25", "30", "35", "30" ] }

[ "D" ]

a contractor undertakes to built a walls in 50 days . he employs 20 peoples for the same . however after 25 days he finds that only 40 % of the work is complete . how many more man need to be employed to complete the work in time ?

20 men complete 0.4 work in 25 days . applying the work rule , m 1 × d 1 × w 2 = m 2 × d 2 × w 1 we have , 20 × 25 × 0.6 = m 2 × 25 × 0.4 or m 2 = 20 × 25 × 0.6 / 25 × 0.4 = 30 men answerd

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "45", "30", "24", "40" ] }

[ "A" ]

the ratio of the capacity to do work of a and b is 3 : 2 . if they together can complete a work in 18 days , then how long does a take to complete the work alone ?

llet a and b take 3 x and 2 x days to complete the work 1 / 3 x + 1 / 2 x = 1 / 18 ⇒ x = 15 so a will take 45 days . answer : a

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "40", "77", "48", "52" ] }

[ "B" ]

in a can , there is a mixture of milk and water in the ratio 4 : 5 . if it is filled with an additional 14 litres of milk the can would be full and ratio of milk and water would become 6 : 5 . find the capacity of the can ?

"let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = 4 / 9 ( t - 14 ) after adding milk , quantity of milk in the mixture = 6 / 11 t . 6 t / 11 - 14 = 4 / 9 ( t - 14 ) 10 t = 1386 - 616 = > t = 77 . answer : b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs 8.82", "rs 9.82", "rs 10.82", "rs 14.82" ] }

[ "D" ]

a fruit seller sells mangoes at the rate of rs . 12 per kg and thereby loses 15 % . at what price per kg , he should have sold them to make a profit of 5 %

"explanation : 85 : 12 = 105 : x x = ( 12 × 105 / 85 ) = rs 14.82 option d"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 40", "rs . 50", "rs . 60", "rs . 70" ] }

[ "C" ]

find the simple interest on rs . 5000 at 6 % per annum for the period from 5 th feb to 19 th april , 2015 .

"explanation : given : 1 ) principal = rs . 5000 2 ) rate of interest = 6 % 3 ) time = 5 th feb to 19 th april , 2015 first find the time period 5 th feb to 19 th april , 2015 feb = 28 – 5 = 23 days march = 31 days april = 19 days total days = 23 + 31 + 19 = 73 days convert days into years , by dividing it by 365 time = 73 / 365 = 1 / 5 simple interest = ( p × r × t ) / 100 = [ 5000 × 6 × ( 1 / 5 ) ] / 100 = rs . 60 simple interest = rs . 60 answer is c"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "105", "30", "35", "20" ] }

[ "A" ]

a contractor undertakes to built a walls in 50 days . he employs 70 peoples for the same . however after 25 days he finds that only 40 % of the work is complete . how many more man need to be employed to complete the work in time ?

"70 men complete 0.4 work in 25 days . applying the work rule , m 1 × d 1 × w 2 = m 2 × d 2 × w 1 we have , 70 × 25 × 0.6 = m 2 × 25 × 0.4 or m 2 = 70 × 25 × 0.6 / 25 × 0.4 = 105 men answera"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "1628.4", "110", "1492", "1496" ] }

[ "B" ]

10 people went to a hotel for combine dinner party 8 of them spent rs . 10 each on their dinner and rest spent 4 more than the average expenditure of all the 10 . what was the total money spent by them .

"solution : let average expenditure of 10 people be x . then , 10 x = 8 * 10 + 2 * ( x + 4 ) ; or , 10 x = 10 * 10 + 2 x + 8 ; or , x = 11 ; so , total money spent = 11 * 10 = rs . 110 . answer : option b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "40.3", "40.4", "40.6", "40.8" ] }

[ "A" ]

the average of 10 numbers is 40.2 . later it is found that two numbers have been wrongly copied . the first is 17 greater than the actual number and the second number added is 13 instead of 31 . find the correct average .

"sum of 10 numbers = 402 corrected sum of 10 numbers = 402 – 13 + 31 – 17 = 403 hence , new average = 403 ⁄ 10 = 40.3 answer a"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 15000", "rs . 15550", "rs . 24750", "rs . 16500" ] }

[ "C" ]

the length of a room is 5.5 m and width is 3.75 m . find the cost of paving the floor by slabs at the rate of rs . 1200 per sq . metre .

"solution area of the floor = ( 5.5 × 3.75 ) m 2 = 20.625 m 2 ∴ cost of paving = rs . ( 1200 × 20.625 ) = 24750 . answer c"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "40", "60", "45", "39" ] }

[ "B" ]

a worker makes a toy in every 2 h . if he works for 120 h , then how many toys will he make ?

"no . of toys = 120 / 2 = 60 answer : b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 15,000", "rs . 15,500", "rs . 15,600", "rs . 20,625" ] }

[ "D" ]

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 . 1000 per sq . metre .

"solution area of the floor = ( 5.5 x 3.75 ) m ² = 20.635 m ² cost of paying = rs . ( 1000 x 20.625 ) = rs . 20625 . answer d"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "16', '", "24', '", "32', '", "48', '" ] }

[ "A" ]

the length and breadth of the floor of the room are 20 feet and 10 feet respectively . square tiles of 2 feet length of different colours are to be laid on the floor . black tiles are laid in the first row on all sides . if white tiles are laid in the one - third of the remaining and blue tiles in the rest , how many blue tiles will be there ?

area left after laying black tiles = [ ( 20 – 4 ) × ( 10 – 4 ) ] sq . ft . = 96 sq . ft . area under white tiles = ( 1 ⁄ 3 × 96 ) sq . ft = 32 sq . ft . area under blue tiles = ( 96 – 32 ) sq . ft = 64 sq . ft . number of blue tiles = 64 / 2 × 2 = 16 . answer a

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs 8.5", "rs 9.5", "rs 10.5", "rs 11.5" ] }

[ "C" ]

a fruit seller sells mangoes at the rate of rs . 8 per kg and thereby loses 20 % . at what price per kg , he should have sold them to make a profit of 5 %

explanation : 80 : 8 = 105 : x x = ( 8 × 105 / 80 ) = rs 10.5 option c

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "12,4", "6,3", "9,3", "6,6" ] }

[ "A" ]

a man can row downstream at the rate of 16 km / hr and upstream at 8 km / hr . find man ' s rate in still water and the rate of current ?

"explanation : rate of still water = 1 / 2 ( 16 + 8 ) = 12 km / hr rate of current = 1 / 2 ( 16 - 8 ) = 4 km / hr answer : option a"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "62.2", "2000", "1700", "1729" ] }

[ "D" ]

1852 - 1230 ÷ 10.00 = ?

"answer given expression = 1852 - 1230 ÷ 10.00 = 1852 - 123 = 1729 correct option : d"

MathQA

{ "label": [ "A", "C", "D", "E" ], "text": [ "124 %", "96 %", "80 %", "64 %" ] }

[ "D" ]

mary ' s income is 60 % more than tim ' s income and tim ' s income is 50 % less than juan ' s income . what % of juan ' s income is mary ' s income .

"even i got 96 % j = 100 t = 100 * 0.5 = 50 m = 50 * 1.6 = 80 if mary ' s income is x percent of j m = j * x / 100 x = m * 100 / j = 80 * 100 / 100 = 80 ans : d"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "1008", "1015", "1022", "1032" ] }

[ "B" ]

the smallest number which when diminished by 7 , is divisible 12 , 16 , 18 , 21 and 28 is :

"explanation required number = ( l . c . m . of 12,16 , 18 , 21 , 28 ) + 7 = 1008 + 7 = 1015 answer b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 1200", "rs . 1300", "rs . 1500", "rs . 4000" ] }

[ "D" ]

a man invests rs . 8000 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 × 5 × 1 ) / 100 = rs . 400 let his additional investment at 8 % = x interest on rs . x at 8 % per annum = ( x × 8 × 1 ) / 100 = 2 x / 25 . to earn 6 % per annum for the total , interest = ( 8000 + x ) × 6 × 1 / 100 . = > 400 + 2 x / 25 = ( 8000 + x ) × 6 × 1 / 100 . = > 40000 + 8 x = ( 8000 + x ) × 6 . = > 40000 + 8 x = 48000 + 6 x . = > 2 x = 8000 . = > x = 4000 . answer : d

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "6693.75", "8032.5", "4462.5", "8900" ] }

[ "A" ]

a sum fetched total simple interest of 4016.25 at the rate of 12 p . c . p . a . in 5 years . what is the sum ?

"let the sums be p . now , 60 % of p = 4016.25 or , p = 6693.75 answer a"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "5.3 %", "7.6 %", "10.6 %", "12 %" ] }

[ "B" ]

a man buys 50 pens at marked price of 46 pens from a whole seller . if he sells these pens giving a discount of 1 % , what is the profit percent ?

"explanation : let marked price be re . 1 each c . p . of 50 pens = rs . 46 s . p . of 50 pens = 99 % of rs . 50 = rs . 49.50 profit % = ( profit / c . p . ) x 100 profit % = ( 3.50 / 46 ) x 100 = 7.6 % answer b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "0.35 %", "0.4 %", "0.045 %", "0.6 %" ] }

[ "C" ]

in expressing a length of 61.472 km as nearly as possible with the 3 significant digits , find the percentage error ?

explanation : error = ( 61.5 - 61.472 ) = 0.028 required percentage = 0.028 / 61.472100 = 0.045 answer : c

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "3 .", "2 .", "1 / 5", "1 / 3 ." ] }

[ "C" ]

two brothers took the gmat exam , the higher score is x and the lower one is y . if the difference between the two scores 1 / 4 , what is the value of y / x ?

"answer is c : 1 / 5 x - y = ( x + y ) / 4 solving for y / x = 1 / 5"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "145', '", "135', '", "75', '", "140', '" ] }

[ "B" ]

if the length of rectangle is three times of its breadth . if the area of rectangle is 6075 sq . m , then calculate the length ?

let breadth = x , length = 3 x . area of rectangle = length * breadth = 3 x ^ 2 = 6075 x ^ 2 = 2025 , x = √ 2025 = 45 m length = 135 m answer b

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "40 m 2", "44 m 2", "52.8 m 2", "36 m 2" ] }

[ "C" ]

the diameter of a garden roller is 1.4 m and it is 2 m long . how much area will it cover in 6 revolutions ? ( use ï € = 22 â „ 7 )

"required area covered in 5 revolutions = 6 ã — 2 ï € rh = 6 ã — 2 ã — 22 â „ 7 ã — 0.7 ã — 2 = 52.8 m 2 answer c"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 7000", "rs . 8000", "rs . 8500", "rs . 9000" ] }

[ "B" ]

the salary of a , b , c , d , e is rs . 8000 , rs . 5000 , rs . 11000 , rs . 7000 , rs . 9000 per month respectively , then the average salary of a , b , c , d , and e per month is

"answer average salary = 8000 + 5000 + 11000 + 7000 + 9000 / 5 = rs . 8000 correct option : b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "$ 21", "$ 27", "$ 31", "$ 43" ] }

[ "D" ]

at an upscale fast - food restaurant , shin can buy 3 burgers , 7 shakes , and one cola for $ 120 . at the same place it would cost $ 158.50 for 4 burgers , 10 shakes , and one cola . how much would it cost for a meal of one burger , one shake , and one cola ?

"let ' s suppose that the price of a burger is bb , of a shake - ss and that of a cola is cc . we can then construct these equations : 3 b + 7 s + c = 120 4 b + 10 s + c = 158.5 subtracting the first equation from the second gives us b + 3 s = 38.5 now if we subtract the new equation two times from first or 3 times from second we will get b + s + c = 43 . in any case , there is no necessity to know each item ' s price , just the sum . answer : d"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 72", "rs . 65", "rs . 54", "rs . 50" ] }

[ "B" ]

the banker ' s gain on a bill due due 1 year hence at 12 % per annum is rs . 7.8 . the true discount is

"solution t . d = [ b . g x 100 / r x t ] = rs . ( 7.8 x 100 / 12 x 1 ) = rs . 65 . answer b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 55.50", "rs . 67.50", "rs . 86.50", "rs . 101.50" ] }

[ "D" ]

the sides of a rectangular field are in the ratio 3 : 4 . if the area of the field is 10092 sq . m , the cost of fencing the field @ 25 paise per metre is

"solution let length = ( 3 x ) metres and breadth = ( 4 x ) metres . then , 3 x × 4 x = 10092 ⇔ 12 x 2 = 10092 ⇔ x 2 = 841 ⇔ x = 29 . so , length = 87 m and breadth = 116 m . perimeter = [ 2 ( 87 + 116 ) ] m = 406 m . ∴ cost of fencing = rs . ( 0.25 × 406 ) = rs . 101.50 . answer d"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 147.50", "rs . 785.50", "rs . 174.50", "rs . 258.50" ] }

[ "C" ]

teas worth rs . 126 per kg and rs . 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs 152 per kg , the price of the third variety per kg will be ?

"explanation : since first and second varieties are mixed in equal proportions . so , their average price = rs . ( 126 + 135 ) / 2 . = > rs . 130.50 . so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . by the rule of alligation , we have : cost of 1 kg cost of 1 kg of 1 st kind of 2 nd kind ( rs . 130.50 ) ( rs . x ) \ / mean price ( rs . 152 ) / \ x â ˆ ’ 152 22.50 = > x â ˆ ’ ( 152 / 22.50 ) = 1 . = > x â ˆ ’ 152 = 22.50 . = > x = 174.50 rs . answer : c"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "11236', '", "11025', '", "14400', '", "12696', '" ] }

[ "A" ]

a lady grows cabbage in her garden that is in the shape of a square . each cabbage takes 1 square foot of area in her garden . this year , she has increased her output by 211 cabbages when compared to last year . the shape of the area used for growing the cabbage has remained a square in both these years . how many cabbages did she produce this year ?

explanatory answer the shape of the area used for growing cabbages has remained a square in both the years . let the side of the square area used for growing cabbages this year be x ft . therefore , the area of the ground used for cultivation this year = x 2 sq . ft . let the side of the square area used for growing cabbages last year be y ft . therefore , the area of the ground used for cultivation last year = y 2 sq . ft . as the number of cabbages grown has increased by 211 , the area would have increased by 211 sq ft because each cabbage takes 1 sq ft space . hence , x 2 - y 2 = 211 ( x + y ) ( x - y ) = 211 . 211 is a prime number and hence it will have only two factors . i . e . , 211 and 1 . therefore , 211 can be expressed as product of 2 numbers in only way = 211 * 1 i . e . , ( x + y ) ( x - y ) = 211 * 1 so , ( x + y ) should be 211 and ( x - y ) should be 1 . solving the two equations we get x = 106 and y = 105 . therefore , number of cabbages produced this year = x 2 = 1062 = 11236 . alternative approach : use answer choices the area in both the years are squares of two numbers . that rules out choice d . 12696 is not the square of any number . check choice a : if this year ' s produce is 11236 , last year ' s produce would have been 11236 - 211 = 11025 11025 is the square of 105 . so , 11236 is the answer . choice a is the correct answer .

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "130 km", "120 km", "100 km", "280 km" ] }

[ "D" ]

a bus covered a distance of 250 km , partly at an average speed of 40 kmph and partly at 60 kmph . if the total time taken is 6.5 hours , then the distance covered at 40 kmph is

"let the partial distance covered at 40 kmph be x let the another partial distance covered at 60 kmph be ( 250 - x ) thus , x / 40 - ( 250 - x ) / 60 = 6.5 or , x / 40 + ( 250 - x ) / 60 = 6.5 or , ( 3 x + 500 - 2 x ) / / 120 = 6.5 or 500 + x = 780 x = 280 answer : d"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "10.22 %", "20.22 %", "21.22 %", "40 %" ] }

[ "D" ]

a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 12 % while buying and by 20 % while selling . what is his percentage profit ?

"the owner buys 100 kg but actually gets 112 kg ; the owner sells 100 kg but actually gives 80 kg ; profit : ( 112 - 80 ) / 80 * 100 = ~ 40 % answer : d ."

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "48", "68", "54", "60" ] }

[ "C" ]

70 % of 300 + 25 % of 400 - ? = 256

"explanation : solution : let 70 % of 300 + 25 % of 400 - x = 256 . then , x = ( 70 / 100 * 300 ) + ( 25 / 100 * 400 ) - 256 = 210 + 100 - 256 = 54 . answer : c"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "95 %", "93 %", "90 %", "80 %" ] }

[ "B" ]

two numbers are respectively 40 % and 30 % more than a third number . the second number expressed in terms of percentage of the first is ?

"here , x = 40 and y = 30 ; therefore second number = [ [ ( 100 + y ) / ( 100 + x ) ] x 100 ] % of first number = [ [ ( 100 + 30 ) / ( 100 + 40 ) ] x 100 ] % of first number = 92.8 % of the first answer : b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "13 , 3", "18 , 4", "15 , 3", "14 , 4" ] }

[ "B" ]

a man can row downstream at 22 kmph and upstream at 14 kmph . find the speed of the man in still water and the speed of stream respectively ?

"explanation : let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 22 - - - ( 1 ) and x - y = 14 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 36 = > x = 18 , y = 4 . answer : option b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "45 ( 4 / 11 ) %", "50 %", "45 ( 5 / 11 ) %", "44 ( 5 / 11 ) %" ] }

[ "B" ]

a batsman scored 120 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 = 120 total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60 total runs scored by running between the wickets = 120 - 60 = 60 required % = ( 60 / 120 ) × 100 = 50 % answer : option b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "25 %", "50 %", "75 %", "95 %" ] }

[ "C" ]

if the radius of a circle is decreased by 50 % , find the percentage decrease in its area .

"let original radius = r . new radius = ( 50 / 100 ) r = ( r / 2 ) original area = 22 / 7 ( r ) 2 = and new area =  ( ( r / 2 ) ) 2 = (  ( r ) 2 ) / 4 decrease in area = ( ( 3 ( 22 / 7 ) ( r ) 2 ) / 4 x ( 1 / ( 22 / 7 ) ( r ) 2 ) x 100 ) % = 75 % answer is c ."

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "0.1602", "0.001602", "1.6021", "0.01602" ] }

[ "D" ]

16.02 × 0.001 = ?

"16.02 × 0.001 = ? or , ? = 0.01602 answer d"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 5725", "rs . 5275", "rs . 6275", "rs . 6500" ] }

[ "D" ]

the owner of a furniture shop charges his customer 30 % more than the cost price . if a customer paid rs . 8450 for a computer table , then what was the cost price of the computer table ?

"cp = sp * ( 100 / ( 100 + profit % ) ) = 8450 ( 100 / 130 ) = rs . 6500 . answer : d"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "10", "5", "9", "15" ] }

[ "C" ]

the sum of the squares of three numbers is 41 , while the sum of their products taken two at a time is 20 . their sum is :

"x ^ + y ^ 2 + z ^ 2 = 138 xy + yz + zx = 131 as we know . . ( x + y + z ) ^ 2 = x ^ 2 + y ^ 2 + z ^ 2 + 2 ( xy + yz + zx ) so ( x + y + z ) ^ 2 = 41 + ( 2 * 20 ) ( x + y + z ) ^ 2 = 81 so x + y + z = 9 answer : c"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 169.50", "rs . 1700", "rs . 195.50", "rs . 180" ] }

[ "C" ]

tea worth rs . 126 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs . 173 per kg , the price of the third variety per kg will be

"solution since first second varieties are mixed in equal proportions , so their average price = rs . ( 126 + 135 / 2 ) = rs . 130.50 so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . x - 173 / 22.50 = 1 = â € º x - 173 = 22.50 = â € º x = 195.50 . hence , price of the third variety = rs . 195.50 per kg . answer c"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "13000", "7000", "10000", "5000" ] }

[ "B" ]

a textile manufacturing firm employees 50 looms . it makes fabrics for a branded company . the aggregate sales value of the output of the 50 looms is rs 5 , 00,000 and the monthly manufacturing expenses is rs 1 , 50,000 . 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 − ( 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 × ( 49 / 50 ) − 150000 × ( 49 / 50 ) − 75000 . = > rs 2 , 68,000 . decrease in profit = > 2 , 75,000 − 2 , 68,000 = > rs . 7,000 . answer : b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "356000", "356500", "357000", "400000" ] }

[ "C" ]

in an election , candidate a got 75 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favour of candidate

"total number of invalid votes = 15 % of 560000 = 15 / 100 × 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 – 84000 = 476000 percentage of votes polled in favour of candidate a = 75 % therefore , the number of valid votes polled in favour of candidate a = 75 % of 476000 = 75 / 100 × 476000 = 35700000 / 100 = 357000 c )"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "4000", "3050", "4400", "4500" ] }

[ "A" ]

if 15 % of 30 % of 50 % of a number is 90 , then what is the number ?

"let the number be a given , 15 / 100 * 30 / 100 * 50 / 100 * a = 90 = > 3 / 20 * 3 / 10 * 1 / 2 * a = 90 = > a = 10 * 20 * 10 * 2 = 4000 . answer : 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 . answer b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "14 / 15", "15 / 14", "4 / 5", "42 / 55" ] }

[ "D" ]

jack and jill are marathon runners . jack can finish a marathon ( 42 km ) in 5.5 hours and jill can run a marathon in 4.2 hours . what is the ratio of their average running speed ? ( jack : jill )

"average speed of jack = distance / time = 42 / ( 11 / 2 ) = 84 / 11 average speed of jill = 42 / ( 4.2 ) = 10 ratio of average speed of jack to jill = ( 84 / 11 ) / 10 = 84 / 110 = 42 / 55 answer d"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "40", "50", "200", "120" ] }

[ "C" ]

the length of a rectangulat plot is 20 metres more than its breadth . if the cost of fencing the plot @ rs . 26.50 per metre is rs . 5300 , what is the length of the plot in metres ?

solution let breadth = x metres then , length = ( x + 20 ) metres . perimeter = ( 5300 / 26.50 ) m = 200 m answer c

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "75", "100", "125", "180" ] }

[ "C" ]

the difference between a number and its three fifth is 50 . what is the number ?

"solution let the number be x . then , x - 3 / 5 x = 50 ‹ = › 2 / 5 x = 50 ‹ = › x = ( 50 x 5 / 2 ) ‹ = › 125 . answer c"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "55", "56", "14", "58" ] }

[ "C" ]

a rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing . if the poles of the fence are kept 20 metres apart , how many poles will be needed ?

"solution perimeter of the plot = 2 ( 90 + 50 ) = 280 m . ∴ number of poles = [ 280 / 20 ] = 14 m answer c"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ ". 0028", ". 0027", ". 0026", ". 0025" ] }

[ "B" ]

what decimal of an hour is a second

explanation : 1 / ( 60 * 60 ) = 1 / 3600 = . 0027 option b

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs 20123.20", "rs 20246.4", "rs 20123.40", "rs 20123.50" ] }

[ "B" ]

what will be the compound interest on rs . 50000 after 3 years at the rate of 12 % per annum

"explanation : ( 50000 × ( 1 + 12 / 100 ) 3 ) = > 50000 × 28 / 25 × 28 / 25 × 28 / 25 = > 70246.4 so compound interest will be 70246.4 - 50000 = rs 20246.4 option b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "243", "253", "312", "432" ] }

[ "B" ]

when a number is divided by 13 , the remainder is 6 . when the same number is divided by 7 , then remainder is 1 . what is the number ?

"explanation : take 243 243 ÷ 7 = 34 , remainder = 5 hence this is not the answer take 312 312 ÷ 7 = 44 , remainder = 4 hence this is not the answer take 253 253 ÷ 7 = 36 , remainder = 1 . 253 ÷ 13 = 19 , remainder = 6 this satisfies both the conditions given in the question . hence it is the answer . answer : b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 15250", "rs . 13375", "rs . 16750", "rs . 18000" ] }

[ "D" ]

a , b and c enter into a partnership . a invests 3 times as much as b invests and 2 / 3 of what c invests . at the end of the year , the profit earned is rs . 66000 . what is the share of c ?

"explanation : let the investment of c be rs . x . the inverstment of b = rs . ( 2 x / 3 ) the inverstment of a = rs . ( 3 × ( 2 / 3 ) x ) = rs . ( 2 x ) ratio of capitals of a , b and c = 2 x : 2 x / 3 : x = 6 : 2 : 3 c ' s share = rs . [ ( 3 / 11 ) × 66000 ] = rs . 18000 answer : option d"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "50.12", "52.12", "51.12", "53.12" ] }

[ "B" ]

a school has 4 section of chemistry in class x having 50 , 35 , 45 and 42 students . the mean marks obtained in chemistry test are 50 , 60 , 55 and 45 respectively for the 4 sections . determine the overall average of marks per student .

"required average marks = 50 ã — 50 + 35 ã — 60 + 45 ã — 55 + 42 ã — 45 / 50 + 35 + 45 + 42 = 2500 + 2100 + 2475 + 1890 / 172 = 8965 â „ 172 = 52.12 answer b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "10 %", "9 %", "11.11 %", "42.85 %" ] }

[ "D" ]

a shopkeeper forced to sell at cost price , uses a 700 grams weight for a kilogram . what is his gain percent ?

"shopkeeper sells 700 g instead of 1000 g . so , his gain = 1000 - 700 = 300 g . thus , % gain = ( 300 * 100 ) / 700 = 42.85 % . answer : option d"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "1", "4", "81", "log 80 243" ] }

[ "B" ]

what is log 3 ( 4 ) log 4 ( 5 ) . . . log 80 ( 81 ) ?

recall the change of base formula logb ( a ) = ln ( a ) ln ( b ) : ( alternatively , we can substitute logc for ln on the right , as long as the base is the same on the top and on the bottom . ) using this , we can rewrite the entire expression using natural logarithms , as ln ( 4 ) ln ( 3 ) ln ( 5 ) ln ( 4 ) ln ( 81 ) ln ( 80 ) = ln ( 81 ) ln ( 3 ) = log 3 ( 81 ) = log 3 ( 34 ) = 4 : correct answer b

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 300", "rs . 200", "rs . 240", "rs . 350" ] }

[ "B" ]

if x gets 25 % more than y and y gets 20 % more than z , the share of z out of rs . 740 will be :

"z share = z , y = 1.2 z x = 1.25 × 1.2 z , x + y + z = 740 ( 1.25 × 1.2 + 1.2 + 1 ) z = 74 3.7 z = 740 , z = 200 answer : . b ."

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 660", "rs . 760", "rs . 600", "rs . 960" ] }

[ "C" ]

a man buys an item at rs . 750 and sells it at the loss of 20 percent . then what is the selling price of that item

"explanation : here always remember , when ever x % loss , it means s . p . = ( 100 - x ) % of c . p when ever x % profit , it means s . p . = ( 100 + x ) % of c . p so here will be ( 100 - x ) % of c . p . = 80 % of 750 = 80 / 100 * 750 = 600 option c"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "20,000", "21,200", "1,800", "20,500" ] }

[ "C" ]

a , b and c start a business each investing 2,000 . after 6 months a withdrew 1000 , b withdrew 1000 and c invests 1000 more . at the end of the year , a total profit of 6600 was recorded . find the share of b .

"ratio of the capitals of a , b and c = 2000 ã — 6 + 1000 ã — 6 : 2000 ã — 6 + 1000 ã — 6 : 2000 ã — 6 + 3000 ã — 6 = 18000 : 18000 : 30000 = 18 : 18 : 30 b â € ™ s share = ( 6600 ã — 18 â „ 66 ) = 1800 answer c"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 4000", "rs . 6000", "rs . 7500", "rs . 6600" ] }

[ "B" ]

3 partners a , b , c starts a business . twice a ' s capital is equal to thrice b ' s capital and b ' s capital is 4 times c ' s capital . out of a total profit of rs . 16500 at the end of the year , b ' s share is

solution let c = x . then , b = 4 x and 2 a = 3 x 4 x = 12 x or a = 6 x . ∴ a : b : c = 6 x : 4 x : x = 6 : 4 : 1 . so b ' s capital = rs ( 16500 x 4 / 11 ) = rs . 6000 . answer b

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "5 kg", "15 kg", "25 kg", "50 kg" ] }

[ "D" ]

the price of rice falls by 20 % . how much rice can be bought now with the money that was sufficient to buy 40 kg of rice previously ?

"solution : let rs . 100 be spend on rice initially for 40 kg . as the price falls by 20 % , new price for 40 kg rice , = ( 100 - 20 % of 100 ) = 80 new price of rice = 80 / 40 = rs . 2 per kg . rice can bought now at = 100 / 2 = 50 kg . answer : option d"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "50", "2.5", "25", ". 25" ] }

[ "A" ]

1 / 0.02 is equal to

explanation : 1 / 0.02 = ( 1 * 100 ) / 2 = 100 / 2 = 50 option a

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "7.5 hr", "8 hr", "8.5 hr", "10 hr" ] }

[ "A" ]

two pipes can fill the cistern in 10 hr and 12 hr respectively , while the third empty it in 20 hr . if all pipes are opened simultaneously , then the cistern will be filled in

"solution : work done by all the tanks working together in 1 hour . 1 / 10 + 1 / 12 − 1 / 20 = 2 / 15 hence , tank will be filled in 15 / 2 = 7.5 hour option ( a )"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "22725", "25675", "22655", "27575" ] }

[ "A" ]

2525 * 9

"explanation : 2525 * ( 10 - 1 ) = 25250 - 2525 = 22725 option a"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 150.40", "rs . 44.80", "rs . 140.40", "rs . 50.48" ] }

[ "B" ]

if the sales tax be reduced from 3 ( 1 / 5 ) % to 2 ( 1 / 2 ) % , then what difference does it make to a person who purchases a bag with marked price of rs . 6400 ?

"explanation : required difference = ( 3 ( 1 / 5 ) of rs . 6400 ) - ( 2 ( 1 / 2 ) of rs . 6400 ) = ( 16 / 5 – 5 / 2 ) % of rs . 6400 = ( 7 / 10 ) x ( 1 / 100 ) x 6400 = rs . 44.80 answer b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "rs . 5870", "rs . 5991", "rs . 6020", "rs . 7670" ] }

[ "D" ]

a grocer has a sale of rs . 5420 , rs . 5660 , rs . 6200 , rs . 6350 and rs . 6500 for 5 consecutive months . find the sale he should have in the sixth month , so that he gets an average sale of rs . 6300 ?

"explanation : total sale for 5 months = rs . ( 5420 + 5660 + 6200 + 6350 + 6500 ) = rs . 30,130 therefore , required sale = rs . [ ( 6300 * 6 ) – 30,130 ] = rs . ( 37800 – 30,130 ) = rs . 7670 answer d"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "60.8", "18", "30", "52.5" ] }

[ "D" ]

a train moves with a speed of 189 kmph . its speed in metres per second is

"solution speed = 189 kmph = ( 189 x 5 / 18 ) m / sec = 52.5 m / sec . answer d"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "148 min .", "150 min .", "145 min .", "155 min ." ] }

[ "B" ]

a man is walking at a speed of 10 km per hour . after every kilometre , he takes rest for 10 minutes . how much time will be take to cover a distance of 10 kilometres ?

"rest time = number of rest ã — time for each rest = 9 ã — 10 = 90 minutes total time to cover 10 km = ( 10 â „ 10 ã — 60 ) minutes + 90 minutes = 150 minutes answer b"

MathQA

{ "label": [ "A", "B", "C", "D" ], "text": [ "111", "1100", "121", "100" ] }

[ "C" ]

find 10 / 11 = 110 / ?

"answer let 10 / 11 = 110 / n , then , 10 n = 11 x 110 ∴ n = ( 11 x 110 ) / 10 = 121 . option : c"

MathQA