regisss/math_qa huggingface dataset:
Problem: a merchant sells an item at a 20 % discount , but still makes a gross profit of 20 percent of the cost . what percent e of the cost would the gross profit on the item have been if it had been sold without the discount ?
Options: a ) 20 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 %
Rationale: "let the market price of the product is mp . let the original cost price of the product is cp . selling price ( discounted price ) = 100 % of mp - 20 % mp = 80 % of mp . - - - - - - - - - - - - - - - - ( 1 ) profit made by selling at discounted price = 20 % of cp - - - - - - - - - - - - - - ( 2 ) apply the formula : profit e = selling price - original cost price = > 20 % of cp = 80 % of mp - 100 % cp = > mp = 120 cp / 80 = 3 / 2 ( cp ) now if product is sold without any discount , then , profit = selling price ( without discount ) - original cost price = market price - original cost price = mp - cp = 3 / 2 cp - cp = 1 / 2 cp = 50 % of cp thus , answer should bec ."
Answer: c
Problem: the width of a rectangle is 10 meter and its area is 150 square meter . if length of the rectangle is increased then its new area is 4 / 3 times of the original area . what is the new perimeter of the rectangle ?
Options: a ) 48 m
b ) 52 m
c ) 54 m
d ) 60 m
e ) none of these
Rationale: let length of new rectangle be x m then x × 10 = 150 × 4 / 3 = 200 or , x = 20 m hence new perimeter = 2 ( 10 + 20 ) = 60 m answer : d
Answer: d
Problem: the cost price of a radio is rs . 2300 and it was sold for rs . 1800 , find the loss % ?
Options: a ) 18 , b ) 99 , c ) 27 , d ) 26 , e ) 21
Rationale: "2300 - - - - 500 100 - - - - ? = > 21 % answer : e"
Answer: e
Problem: a and b can finish a work 30 days if they work together . they worked together for 20 days and then b left . a finished the remaining work in another 20 days . in how many days a alone can finish the work ?
Options: a ) 30 , b ) 60 , c ) 40 , d ) 20 , e ) 50
Rationale: amount of work done by a and b in 1 day = 1 / 30 amount of work done by a and b in 20 days = 20 × ( 1 / 30 ) = 20 / 30 = 2 / 3 remaining work – 1 – 2 / 3 = 1 / 3 a completes 1 / 3 work in 20 days amount of work a can do in 1 day = ( 1 / 3 ) / 20 = 1 / 60 = > a can complete the work in 60 days answer is b .
Answer: b
Problem: 0.0004 ? = 0.01
Options: a ) 4 , b ) 0.4 , c ) 0.04 , d ) 40 , e ) none of these
Rationale: "explanation : required answer = 0.0004 / 0.01 = 0.04 / 1 = 0.04 . answer : option c"
Answer: c
Problem: 860 % of 50 + 50 % of 860 = ?
Options: a ) 430 , b ) 516 , c ) 660 , d ) 860 , e ) none
Rationale: "solution given expression = [ 860 / 100 x 50 + 50 / 100 x 860 ] = 430 + 430 = 860 . answer d"
Answer: d
Problem: in a lake , there is a patch of lily pads . every day , the patch doubles in size . it takes 39 days for the patch to cover the entire lake , how many days would it take the patch to cover half of the lake ?
Options: a ) 36 , b ) 2 ^ 4 * 3 , c ) 24 , d ) 38 , e ) 47
Rationale: "working backward from the day it ' s covered : day 39 : fully covered day 38 : half covered so 38 days answer : d"
Answer: d
Problem: find the product of the localvalue and absolutevalue of 4 in 20468 ?
Options: a ) 1000 , b ) 1100 , c ) 1200 , d ) 1300 , e ) 1600
Rationale: "local value of 4 = 4 x 100 = 400 place value of 4 = 4 there fore = 4 x 400 = 1600 e"
Answer: e
Problem: marcella has 20 pairs of shoes . if she loses 9 individual shoes , what is the greatest number of matching pairs she could have left ?
Options: a ) 21 , b ) 20 , c ) 19 , d ) 16 , e ) 15
Rationale: "marcella has 20 pairs of shoes and loses 9 shoes . to minimize the loss of identical pairs of shoes we want marcella to lose as many identical pairs as possible . this would yield 4 identical pairs and 1 additional shoe ( destroying 5 pairs of shoes ) . the 20 pairs of shoes minus the 5 ' destroyed ' pairs yields 15 pairs that still fulfill the requirements . answer : e"
Answer: e
Problem: a miniature roulette wheel is divided into 10 equal sectors , each bearing a distinct integer from 1 to 10 , inclusive . each time the wheel is spun , a ball randomly determines the winning sector by settling in that sector . if the wheel is spun five times , approximately what is the probability that the product of the five winning sectors ’ integers will be even ?
Options: a ) 50 % , b ) 67 % , c ) 85 % , d ) 91 % , e ) 97 %
Rationale: "the only way to have an odd product is if all 5 integers are odd . p ( odd product ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 32 p ( even product ) = 1 - 1 / 32 = 31 / 32 which is about 97 % the answer is e ."
Answer: e
Problem: on a map , 7 centimeters represents 35 kilometers . two cities 245 kilometers apart would be separated on the map by how many centimeters ?
Options: a ) 15 , b ) 37 , c ) 49 , d ) 110 , e ) 180
Rationale: 1 centimeter represents 5 kilometers ( 35 / 7 ) x = 245 / 5 = 49 answer : c
Answer: c
Problem: a certain bag contains 60 balls — 22 white , 10 green , 7 yellow , 15 red , and 6 purple . if a ball is to be chosen at random , what is the probability that the ball will be neither red nor purple ?
Options: a ) 0.09 , b ) 0.65 , c ) 0.54 , d ) 0.85 , e ) 0.91
Rationale: "according to the stem the ball can be white , green or yellow , so the probability is ( white + green + yellow ) / ( total ) = ( 22 + 10 + 17 ) / 60 = 39 / 60 = 0.65 . answer : b ."
Answer: b
Problem: points a , b , c , and d , in that order , lie on a line . if ab = 2 cm , ac = 5 cm , and bd = 9 cm , what is cd , in centimeters ?
Options: a ) 1 , b ) 2 , c ) 3 , d ) 6 , e ) 5
Rationale: "putting a value to each point , lets use the following : a - 0 b - 2 ( ab = 2 ) c - 5 ( ac = 5 ) d - 11 ( bd = 9 ) cd is 11 - 5 = 6 . ans d"
Answer: d
Problem: if the area of a square with sides of length 4 centimeters is equal to the area of a rectangle with a width of 8 centimeters , what is the length of the rectangle , in centimeters ?
Options: a ) 4 , b ) 8 , c ) 2 , d ) 16 , e ) 18
Rationale: "let length of rectangle = l 4 ^ 2 = l * 8 = > l = 16 / 8 = 2 answer c"
Answer: c
Problem: the average of 6 numbers is 6.8 . if one of the numbers is multiplied by a factor of 3 , the average of the numbers increases to 9.2 . what number is multiplied by 3 ?
Options: a ) 9.5 , b ) 8.0 , c ) 7.2 , d ) 5.0 , e ) 4.0
Rationale: "the average of 6 numbers is 6.8 the sum of 6 numbers will be 6.8 x 6 = 40.8 the average of 6 number after one of the number is multiplied by 3 is 9.2 the sum of the numbers will now be 9.2 x 6 = 55.2 so the sum has increased by 55.2 - 40.8 = 14.4 let the number multiplied by 3 be n then , 3 n = n + 14.4 or 2 n = 14.4 or n = 7.2 answer : - c"
Answer: c
Problem: two trains 200 m and 800 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ?
Options: a ) 10.6 , b ) 10.9 , c ) 36 , d ) 10.8 , e ) 10.1
Rationale: "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 200 + 800 = 1000 m . required time = 1000 * 9 / 250 = 36 sec . answer : c"
Answer: c
Problem: if 3 < x < 6 < y < 7 , then what is the greatest possible positive integer difference of x and y ?
Options: a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
Rationale: "3 < x < 6 < y < 7 ; 3 < x y < 7 3 + y < x + 7 y - x < 4 . positive integer difference is 3 ( for example y = 6.5 and x = 3.5 ) answer : a ."
Answer: a
Problem: the ratio between the perimeter and the width of a rectangle is 5 : 1 . if the area of the rectangle is 96 square centimeters , what is the length of the rectangle in centimeters ?
Options: a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15
Rationale: "perimeter = 2 ( w + l ) = 5 w 3 w = 2 l w = 2 l / 3 wl = 96 2 l ^ 2 / 3 = 96 l ^ 2 = 144 l = 12 cm the answer is b ."
Answer: b
Problem: in a group of 90 people , 55 have visited iceland and 33 have visited norway . if 51 people have visited both iceland and norway , how many people have visited neither country ?
Options: a ) 50 , b ) 51 , c ) 52 , d ) 53 , e ) 54
Rationale: "this is an example of a standard overlapping sets question . it has no ' twists ' to it , so you ' ll likely find using the overlapping sets formula to be a fairly easy approach . if you ' re not familiar with it , then here is the formula : 90 = 55 + 33 - 51 + ( # in neither group ) = 53 the prompt gives you all of the numbers you need to get to the correct answer . just plug in and solve . d"
Answer: d
Problem: a small company reduced its faculty by approximately 14 percent to 195 employees . what was the original number of employees ?
Options: a ) a ) 182 , b ) b ) 227 , c ) c ) 220 , d ) d ) 224 , e ) e ) 302
Rationale: "if x is the original number of employees , then after 14 % reduction in employees number is . 86 x but we are given . 86 x = 195 x = 227 so the original number of employees is 227 correct answer - b"
Answer: b
Problem: in a dairy farm , 40 cows eat 40 bags of husk in 40 days . in how many days one cow will eat one bag of husk ?
Options: a ) 1 , b ) 1 / 40 , c ) 40 , d ) 80 , e ) 1 / 80
Rationale: "assume that in x days , one cow will eat one bag of husk . more cows , less days ( indirect proportion ) more bags , more days ( direct proportion ) hence we can write as cowsbags 40 : 11 : 40 ⎫ ⎭ ⎬ : : x : 40 ⇒ 40 × 1 × 40 = 1 × 40 × x ⇒ x = 40 answer : c"
Answer: c
Problem: the average age of 4 men is increased by years when two of them whose ages are 21 years and 23 years are replaced by two new men . the average age of the two new men is
Options: a ) 22 , b ) 30 , c ) 99 , d ) 26 , e ) 27
Rationale: total age increased = ( 8 * 2 ) years = 8 years . sum of ages of two new men = ( 21 + 23 + 8 ) years = 52 years average age of two new men = ( 52 / 2 ) years = 26 years . answer : d
Answer: d
Problem: one type of liquid contains 25 % of kerosene , the other contains 30 % of kerosene . p can is filled with 6 parts of the first liquid and 4 parts of the second liquid . find the percentage of the kerosene in the new mixture .
Options: a ) 28 % , b ) 25 % , c ) 30 % , d ) 27 % , e ) 35 %
Rationale: let p be filled by 60 lts of 1 st liquid and 40 lts . of 2 nd liquid . amount of kerosene = ( 25 * 60 / 100 ) + ( 30 * 40 / 100 ) = 27 lts . % of kerosene = 27 % answer : d
Answer: d
Problem: the ratio of income of anand to that of balu is 5 : 4 and the expenditure of anand to that of balu is 3 : 2 . if at the end of the year , each saves rs , 800 , the income of anand is :
Options: a ) rs 1600 , b ) rs 1800 , c ) rs 2000 , d ) rs 2200 , e ) none of these
Rationale: explanation : let the income of anand and balu be 5 x and 4 x and . the expenditures of anand and balu be 3 y and 2 y . then , 5 x 3 y = 800 and 4 x 2 y = 800 . on solving we get : x = 400 . as income = 5 x = rs . 2000 . answer : c
Answer: c
Problem: the largest natural number by which the product of 3 consecutive even natural numbers is always divisible , is
Options: a ) 16 , b ) 24 , c ) 48 , d ) 96 , e ) 78
Rationale: solution required number = ( 2 × 4 × 6 ) = 48 . answer c
Answer: c
Problem: a sum of money is put out at compound interest for 2 years at 20 % . it would fetch rs . 482 more if the interest were payable half - yearly , then it were pay able yearly . find the sum ?
Options: a ) 2099 , b ) 6086 , c ) 2976 , d ) 2000 , e ) 3862
Rationale: "p ( 11 / 10 ) 4 - p ( 6 / 5 ) 2 = 482 p = 2000 answer : d"
Answer: d
Problem: a group of hikers is planning a trip that will take them up a mountain using one route and back down using another route . they plan to travel down the mountain at a rate of one and a half times the rate they will use on the way up , but the time each route will take is the same . if they will go up the mountain at a rate of 5 miles per day and it will take them two days , how many miles long is the route down the mountain ?
Options: a ) 4 , b ) 6 , c ) 8 , d ) 12 , e ) 15
Rationale: "on the way down , the rate is 1.5 * 5 = 7.5 miles per day . the distance of the route down the mountain is 2 * 7.5 = 15 miles . the answer is e ."
Answer: e
Problem: in the xy - coordinate plane , the graph of y = - x ^ 2 + 4 intersects line l at ( p , 5 ) and ( t , - 7 ) . what is the least possible value of the slope of line l ?
Options: a ) 6 , b ) 2 , c ) - 4 , d ) - 6 , e ) - 10
Rationale: "we need to find out the value of p and l to get to the slope . line l and graph y intersect at point ( p , 5 ) . hence , x = p and y = 5 should sactisfy the graph . soliving 5 = - p 2 + 4 p 2 = 1 p = + or - 1 simillarly point ( t , - 7 ) should satisfy the equation . hence x = t and y = - 7 . - 7 = - t 2 + 9 t = + or - 4 considering p = - 2 and t = 4 , the least slope is ( - 7 - 5 ) / ( 4 - 1 ) = - 4 imo option c is correct answer ."
Answer: c
Problem: the value of a machine depreciates at 25 % per annum . if its present value is $ 1 , 50,000 , at what price should it be sold after two years such that a profit of $ 24,000 is made ?
Options: a ) $ 258375 , b ) $ 438375 , c ) $ 128375 , d ) $ 108375 , e ) $ 158375
Rationale: "the value of the machine after two years = 0.75 * 0.75 * 1 , 50,000 = $ 84,375 sp such that a profit of $ 24,000 is made = 84,375 + 24,000 = $ 1 , 08,375 d"
Answer: d
Problem: triangle atriangle b are similar triangles with areas 2268 units square and 2527 units square respectively . the ratio of there corresponding height would be
Options: a ) 9 : 10 , b ) 18 : 19 , c ) 23 : 27 , d ) 13 : 17 , e ) 15 : 23
Rationale: "let x be the height of triangle a and y be the height of triangle of b . since triangles are similar , ratio of area of a and b is in the ratio of x ^ 2 / y ^ 2 therefore , ( x ^ 2 / y ^ 2 ) = 2268 / 2527 ( x ^ 2 / y ^ 2 ) = ( 18 * 18 * 7 ) / ( 19 * 19 * 7 ) ( x ^ 2 / y ^ 2 ) = 18 ^ 2 / 19 ^ 2 x / y = 18 / 19 ans = b"
Answer: b
Problem: 40 litres of diesel is required to travel 600 km using a 800 cc engine . if the volume of diesel required to cover a distance varies directly as the capacity of the engine , then how many litres of diesel is required to travel 800 km using 1200 cc engine ?
Options: a ) 80 litres , b ) 90 litres , c ) 120 litres , d ) 170 litres , e ) none of these
Rationale: "explanatory answer to cover a distance of 800 kms using a 800 cc engine , the amount of diesel required = 800 / 600 * 40 = 53.33 litres . however , the vehicle uses a 1200 cc engine and the question states that the amount of diesel required varies directly as the engine capacity . i . e . , for instance , if the capacity of engine doubles , the diesel requirement will double too . therefore , with a 1200 cc engine , quantity of diesel required = 1200 / 800 * 53.33 = 80 litres . answer a"
Answer: a
Problem: a and b go around a circular track of length 600 m on a cycle at speeds of 24 kmph and 60 kmph . after how much time will they meet for the first time at the starting point ?
Options: a ) 120 sec , b ) 176 sec , c ) 178 sec , d ) 180 sec , e ) 189 sec
Rationale: "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 600 / ( 24 * 5 / 18 ) , 600 / ( 60 * 5 / 18 ) } = 180 sec . answer : d"
Answer: d
Problem: the captain of a cricket team of 11 members is 28 years old and the wicket keeper is 3 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ?
Options: a ) 20 years , b ) 21 years , c ) 22 years , d ) 23 years , e ) 25 years
Rationale: "explanation : let the average age of the whole team by x years . 11 x â € “ ( 28 + 31 ) = 9 ( x - 1 ) 11 x â € “ 9 x = 50 2 x = 50 x = 25 so , average age of the team is 25 years . answer e"
Answer: e
Problem: during 2003 , a company produced an average of 2,000 products per month . how many products will the company need to produce from 2004 through 2007 in order to increase its monthly average for the period from 2003 through 2007 by 150 % over its 2003 average ?
Options: a ) 287,000 , b ) 290,000 , c ) 284,000 , d ) 285,000 , e ) 286,000
Rationale: company produced 12 * 2000 = 24,000 products in 2003 . if company produces x products from 2004 to 2007 , then total amount of product produced in 4 years ( 2003 through 2007 ) is x + 24,000 . the gives the average of ( x + 24,000 ) / 4 . this average needs to be 300 % higher than that in 2003 . in math terms , 36,000 + 150 % ( 36,000 ) = 80,000 . so : ( x + 36,000 ) / 4 = 80,000 x + 36,000 = 320,000 x = 284,000 the answer is c .
Answer: c
Problem: a library has an average of 510 visitors on sundays and 240 on other day . the average number of visitors in a month of 30 days starting with sunday is
Options: a ) 280 , b ) 285 , c ) 290 , d ) 295 , e ) 275
Rationale: "explanation : as the month begin with sunday , so there will be five sundays in the month . so result will be : = ( 510 × 5 + 240 × 25 / 30 ) = ( 8550 / 30 ) = 285 option b"
Answer: b
Problem: a train speeds past a pole in 15 sec and a platform 150 m long in 25 sec , its length is ?
Options: a ) 100 , b ) 150 , c ) 160 , d ) 225 , e ) 180
Rationale: "let the length of the train be x m and its speed be y m / sec . then , x / y = 15 = > y = x / 15 ( x + 150 ) / 25 = x / 15 = > x = 225 m . answer : option d"
Answer: d
Problem: there are 2 sections a and b in a class , consisting of 40 and 30 students respectively . if the average weight of section a is 50 kg and that of section b is 60 kg , find the average of the whole class ?
Options: a ) 50.78 kg , b ) 49.32 kg , c ) 61.67 kg , d ) 69.15 kg , e ) 54.28 kg
Rationale: "total weight of 40 + 30 students = 40 * 50 + 30 * 60 = 2000 + 1800 average weight of the class is = 3800 / 70 = 54.28 kg answer is e"
Answer: e
Problem: a customer purchased a package of ground beef at a cost of $ 1.76 per pound . for the same amount of money , the customer could have purchased a piece of steak that weighed 36 percent less than the package of ground beef . what was the cost per pound of the steak ?
Options: a ) $ 2.00 , b ) $ 2.25 , c ) $ 2.50 , d ) $ 2.75 , e ) $ 3.00
Rationale: "for simplicity , let ' s assume the customer bought 1 pound of ground beef for $ 1.76 . let x be the price per pound for the steak . then 0.64 x = 176 x = 176 / 0.64 = $ 2.75 the answer is d ."
Answer: d
Problem: a train 275 meters long is running with a speed of 60 kmph . in what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going ?
Options: a ) 7 sec , b ) 5 sec , c ) 15 sec , d ) 6 sec , e ) 12 sec
Rationale: "speed of train relative to man = ( 60 + 6 ) km / hr = 66 km / hr [ 66 * 5 / 18 ] m / sec = [ 55 / 3 ] m / sec . time taken to pass the man = [ 275 * 3 / 55 ] sec = 15 sec answer : c"
Answer: c
Problem: lagaan is levied on the 60 percent of the cultivated land . the revenue department collected total rs . 3 , 74,000 through the lagaan from the village of mutter . mutter , a very rich farmer , paid only rs . 480 as lagaan . the percentage of total land of mutter over the total taxable land of the village is :
Options: a ) 0.20833 , b ) 0.14544 , c ) 0.25632 , d ) 0.2139 , e ) 0.63435
Rationale: "total land of sukhiya = \ inline \ frac { 480 x } { 0.6 } = 800 x \ therefore cultivated land of village = 374000 x \ therefore required percentage = \ inline \ frac { 800 x } { 374000 } \ times 100 = 0.21390 d"
Answer: d
Problem: the average ( arithmetic mean ) of the even integers from 16 to 44 inclusive is how much greater than the average ( arithmetic mean ) of the even integers from 14 to 56 inclusive ?
Options: a ) 10 , b ) 5 , c ) 15 , d ) 20 , e ) 7
Rationale: "so , according to a mean of a set of even numbers from 16 to 44 = ( 16 + 44 ) / 2 = 30 and mean of a set of even numbers from 14 to 56 = ( 14 + 56 ) / 2 = 35 difference = 35 - 30 = 5 answer : b ."
Answer: b
Problem: if a card is drawn from a well shuffled pack of cards , the probability of drawing a spade or a king is ?
Options: a ) 4 / 15 , b ) 4 / 10 , c ) 4 / 18 , d ) 4 / 13 , e ) 4 / 11
Rationale: "p ( s ᴜ k ) = p ( s ) + p ( k ) - p ( s ∩ k ) , where s denotes spade and k denotes king . p ( s ᴜ k ) = 13 / 52 + 4 / 52 - 1 / 52 = 4 / 13 answer : d"
Answer: d
Problem: two employees a and b are paid a total of rs . 570 per week by their employer . if a is paid 150 percent of the sum paid to b , how much is b paid per week ?
Options: a ) s . 130 , b ) s . 140 , c ) s . 150 , d ) s . 228 , e ) s . 282
Rationale: "let the amount paid to a per week = x and the amount paid to b per week = y then x + y = 570 but x = 150 % of y = 150 y / 100 = 15 y / 10 ∴ 15 y / 10 + y = 570 ⇒ y [ 15 / 10 + 1 ] = 570 ⇒ 25 y / 10 = 570 ⇒ 25 y = 5700 ⇒ y = 5700 / 25 = rs . 228 d )"
Answer: d
Problem: a company has two models of computers , model m and model n . operating at a constant rate , a model m computer can complete a certain task in 36 minutes and a model n computer can complete the same task in 18 minutes . if the company used the same number of each model of computer to complete the task in 1 minute , how many model m computers were used ?
Options: a ) 12 , b ) 11 , c ) 10 , d ) 9 , e ) 8
Rationale: "let ' s say 1 work is processing 36 gb of data . model m : 1 gb per min model n : 2 gb per min working together , 1 m and 1 n = 3 gb per min so , 12 times as many computers would work at 36 gb per min . so no . of m = 12 answer is a"
Answer: a
Problem: a retailer sold an appliance for $ 70 . if the retailer ' s gross profit on the appliance was 25 percent of the retailer ' s cost for the appliance , how many dollars was the retailer ' s gross profit ?
Options: a ) $ 14 , b ) $ 16 , c ) $ 20 , d ) $ 24 , e ) $ 25
Rationale: "gross profit = selling price - cost = 70 - x 70 - x = 1 / 4 x x = 56 gp = 70 - 56 = 14"
Answer: a
Problem: a sporting goods store sold 60 frisbees in one week , some for $ 3 and the rest for $ 4 each . if receipts from frisbee sales for the week totaled $ 200 , what is the fewest number of $ 4 frisbees that could have been sold ?
Options: a ) 24 , b ) 12 , c ) 20 , d ) 4 , e ) 2
Rationale: the wording of the question is slightly misleading . when a question asks what is thefewestthatcouldhave been sold , it is suggesting that there could be many different quantities sold . in this question however , because we are told that exactly 64 frisbees have been sold and revenue was exactly $ 204 , there is only one possible solution for the number of $ 3 and $ 4 frisbees sold . to solve , we have 2 equations and 2 unknowns let x = number of $ 3 frisbees sold let y = number of $ 4 frisbees sold x + y = 60 3 x + 4 y = 200 x = 60 - y 3 ( 60 - y ) + 4 y = 200 180 - 3 y + 4 y = 200 y = 20 answer : c
Answer: c
Problem: there are 16 bees in the hive , then 7 more fly . how many bees are there in all ?
Options: a ) 7 , b ) 23 , c ) 12 , d ) 17 , e ) 25
Rationale: 16 + 7 = 23 . answer is b .
Answer: b
Problem: laura took out a charge account at the general store and agreed to pay 6 % simple annual interest . if she charges $ 35 on her account in january , how much will she owe a year later , assuming she does not make any additional charges or payments ?
Options: a ) $ 2.10 , b ) $ 37.10 , c ) $ 37.16 , d ) $ 38.10 , e ) $ 38.80
Rationale: "principal that is amount taken by laura at year beginning = 35 $ rate of interest = 6 % interest = ( 6 / 100 ) * 35 = 2.10 $ total amount that laura owes a year later = 35 + 2.1 = 37.1 $ answer b"
Answer: b
Problem: with both inlets open , a water tank will be filled with water in 24 minutes . the first inlet alone would fill the tank in 2 hours . if in every minutes the second inlet admits 50 cubic meters of water than the first , what is the capacity of the tank ?
Options: a ) 2,000 , b ) 10,500 , c ) 11,750 , d ) 12,000 , e ) 13,000
Rationale: the work done by inlet a and b together in 1 min = 1 / 24 the work done by inlet a ( first inlet ) in 1 min = 1 / 120 the work done by inlet b ( second inlet ) in 1 min = ( 1 / 24 ) - ( 1 / 120 ) = 1 / 30 difference of work done by b and a = b - a = 50 cubic meter i . e . ( 1 / 30 ) - ( 1 / 120 ) = 50 cubic meter i . e . tank = 40 * 50 = 2000 cubic meter answer : option a
Answer: a
Problem: if 4 men can colour 48 m long cloth in 2 days , then 6 men can colour 36 m long cloth in ?
Options: a ) 1 day , b ) 2 days , c ) 3 days , d ) 4 days , e ) 5 days
Rationale: "the length of cloth painted by one man in one day = 48 / 4 × 2 = 6 m no . of days required to paint 36 m cloth by 6 men = 36 / 6 × 6 = 1 day . option ' a '"
Answer: a
Problem: find the area of a rhombus one side of which measures 20 cm and one diagonal is 23 cm .
Options: a ) 100 cm 2 , b ) 150 cm 2 , c ) 300 cm 2 , d ) 368 cm 2 , e ) 400 cm 2
Rationale: "explanation : let other diagonal = 2 x cm . since diagonals of a rhombus bisect each other at right angles , we have : ( 20 ) 2 = ( 12 ) 2 + ( x ) 2 = > x = √ ( 20 ) 2 – ( 12 ) 2 = √ 256 = 16 cm . _ i so , other diagonal = 32 cm . area of rhombus = ( 1 / 2 ) x ( product of diagonals ) = ( 1 / 2 × 23 x 32 ) cm 2 = 368 cm 2 answer : option d"
Answer: d
Problem: compound interest of rs . 6000 at 10 % per annum for 1 1 / 2 years will be ( interest compounded half yearly ) .
Options: a ) rs . 473 , b ) rs . 374 , c ) rs . 495 , d ) rs . 945 , e ) none of the above
Rationale: "10 % interest per annum will be 5 % interest half yearly for 3 terms ( 1 1 / 2 years ) so compound interest = 6000 [ 1 + ( 5 / 100 ) ] ^ 3 - 6000 = 6000 [ ( 21 / 20 ) ^ 3 - 1 ] = 6000 ( 9261 - 8000 ) / 8000 = 6 * 1261 / 8 = 945 answer : d"
Answer: d
Problem: a and b started a business investing rs . 75,000 and rs 20,000 respectively . in what ratio the profit earned after 2 years be divided between a and b respectively ?
Options: a ) 9 : 2 , b ) 3 : 2 , c ) 18 : 20 , d ) 18 : 4 , e ) 13 : 4
Rationale: "a : b = 75000 : 20000 = 75 : 20 = 13 : 4 answer : e"
Answer: e
Problem: a person borrows 5000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 61 ⁄ 4 % p . a . for 2 years . find his gain in the transaction per year .
Options: a ) 112.50 , b ) 125 , c ) 150 , d ) 167.5 , e ) none of these
Rationale: "gain in 2 years = [ ( 5000 × 25 / 4 × 2 / 100 ) − ( 5000 × 4 × 2 / 100 ) ] = ( 625 – 400 ) = 225 . ∴ gain in 1 year = ( 225 ⁄ 2 ) = 112.50 answer a"
Answer: a
Problem: a side of beef lost 30 percent of its weight in processing . if the side of beef weighed 500 pounds after processing , how many pounds did it weigh before processing ?
Options: a ) 714.28 , b ) 712 , c ) 714.88 , d ) 714.25 , e ) 714
Rationale: "let weight of side of beef before processing = x ( 70 / 100 ) * x = 500 = > x = ( 500 * 100 ) / 70 = 714.28 answer a"
Answer: a
Problem: if the price of a tv is first decreased by 20 % and then increased by 40 % , then the net change in the price will be :
Options: a ) 4 % increase , b ) 12 % increase , c ) 10 % decrease , d ) 6 % increase , e ) none of these
Rationale: "explanation : solution : let the original price be rs . 100 . new final price = 140 % of ( 80 % of 100 ) = rs . 140 / 100 * 80 / 100 * 100 = rs . 112 . . ' . increase = 12 % answer : b"
Answer: b
Problem: two sets of 3 consecutive positive odd integers have exactly one integer in common . the sum of the integers in the set with greater numbers is how much greater than the sum of the integers in the other set ?
Options: a ) 4 , b ) 7 , c ) 8 , d ) 12 , e ) it can not be determined from the information given .
Rationale: a = ( 1 , 3,5 ) , sum of this = 9 b = ( 5 , 7,9 ) , sum of this = 21 , the differenct between 21 - 9 = 12 hence , 12 is the answer i . e . d
Answer: d
Problem: a train running at 1 / 5 of its own speed reached a place in 20 hours . how much time could be saved if the train would have run at its own speed ?
Options: a ) 8 hrs , b ) 10 hrs , c ) 12 hrs , d ) 16 hrs , e ) 6 hrs
Rationale: "time taken if run its own speed = 1 / 5 * 20 = 4 hrs time saved = 20 - 4 = 16 hrs answer : d"
Answer: d
Problem: in a certain pond , 60 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the pond ?
Options: a ) 400 , b ) 1,500 , c ) 1,250 , d ) 2,500 , e ) 10,000
Rationale: "this is a rather straight forward ratio problem . 1 . 60 fish tagged 2 . 2 out of the 50 fish caught were tagged thus 2 / 50 2 / 50 = 60 / x thus , x = 1500 think of the analogy : 2 fish is to 50 fish as 50 fish is to . . . ? you ' ve tagged 50 fish and you need to find what that comprises as a percentage of the total fish population - we have that information with the ratio of the second catch . b"
Answer: b
Problem: how many 4 - digit even numbers are possible such that if one of the digits is 5 , the next / succeeding digit to it should be 7 ?
Options: a ) 5 , b ) 305 , c ) 365 , d ) 50 , e ) 495
Rationale: 5700 , 5710 , . . . . . . 5798 , so total 50 . hence option d .
Answer: d
Problem: a man has 1044 candles . after burning , he can make a new candle from 9 stubs left behind . find the maximum number of candles that can be made .
Options: a ) 116 , b ) 120 , c ) 130 , d ) 140 , e ) none
Rationale: explanation : since , 1 candle is made from 9 stubs . therefore , from 1044 stubs , 1044 / 9 i . e 116 candles will be made . now , from 116 stubs , 116 / 9 i . e 12 candles can be made and 8 stubs will be left . thus , total stubs left is ( 12 + 8 ) = 20 out of which 20 / 9 i . e 2 candles can be made and 2 stubs will be left . thus , maximum number of candles that can be made is 130 ( 116 + 12 + 2 ) . answer : c
Answer: c
Problem: find the amount on rs . 7000 in 2 years , the rate of interest being 4 % per first year and 5 % for the second year ?
Options: a ) 5568 , b ) 6369 , c ) 5460 , d ) 7644 , e ) 6734
Rationale: "7000 * 104 / 100 * 105 / 100 = > 7644 answer : d"
Answer: d
Problem: a merchant has 1000 kg of sugar part of which he sells at 8 % profit and the rest at 18 % profit . he gains 14 % on the whole . the quantity sold at 18 % profit is
Options: a ) 300 , b ) 400 , c ) 600 , d ) 500 , e ) none of these
Rationale: explanation : by the rule of alligation , we have profit % by selling 1 st part profit % by selling 2 nd part 8 18 net % profit 14 18 - 14 = 4 14 - 8 = 6 = > quantity of part 1 : quantity of part 2 = 4 : 6 = 2 : 3 total quantity is given as 1000 kg so quantity of part 2 ( quantity sold at 18 % profit ) = 1000 × 35 = 600 kg answer : option c
Answer: c
Problem: a trader sells 40 metres of cloth for rs . 8200 at a profit of rs . 25 per metre of cloth . how much profit will the trder earn on 40 metres of cloth ?
Options: a ) rs . 950 , b ) rs . 1500 , c ) rs . 1000 , d ) rs . 1200 , e ) none of these
Rationale: "explanation : sp of 1 metre cloth = 8200 / 40 = rs . 205 . cp of 1 metre cloth = rs . 205 – 25 = rs . 180 cp on 40 metres = 180 x 40 = rs . 7200 profit earned on 40 metres cloth = rs . 8200 – rs . 7200 = rs . 1000 . answer : option c"
Answer: c
Problem: a batsman makes a score of 64 runs in the 16 th innings and thus increased his average by 3 . find his average after the 16 th inning ?
Options: a ) 23 , b ) 28 , c ) 27 , d ) 19 , e ) 24
Rationale: "let the average after the 16 th inning be p . so , the average after the 15 th inning will be ( p - 3 ) hence , 15 ( p - 30 ) + 64 = 16 p = > p = 19 . answer : d"
Answer: d
Problem: when 2 is added to half of one - third of one - fifth of a number , the result is one - fifteenth of the number . find the number ?
Options: a ) 40 , b ) 65 , c ) 73 , d ) 60 , e ) 80
Rationale: "d 60 let the number be 2 + 1 / 2 [ 1 / 3 ( a / 5 ) ] = a / 15 = > 2 = a / 30 = > a = 60"
Answer: d
Problem: a sells a bicycle to b at a profit of 20 % . b sells it to c at a profit of 25 % . if c pays rs . 225 for it , the cost price of the bicycle for a is
Options: a ) 627 , b ) 297 , c ) 127 , d ) 150 , e ) 181
Rationale: explanation : 125 % of 120 % of a = 225 125 / 100 * 120 / 100 * a = 225 a = 225 * 2 / 3 = 150 . answer : d
Answer: d
Problem: the perimeter of a triangle is 32 cm and the inradius of the triangle is 3.5 cm . what is the area of the triangle ?
Options: a ) 22 , b ) 35 , c ) 56 , d ) 54 , e ) 23
Rationale: "area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 3.5 * 32 / 2 = 56 cm 2 answer : c"
Answer: c
Problem: if circles x and y have the same area and circle x has a circumference of 10 π , half of the radius of circle y is :
Options: a ) 16 , b ) 8 , c ) 4 , d ) 2.5 , e ) 1
Rationale: "x be radius of circle x y be radius of circle y given : pi * x ^ 2 = pi * y ^ 2 also , 2 * pi * x = 10 * pi x = 5 thus y = 5 y / 2 = 2.5 ans : d"
Answer: d
Problem: how many positive integers , from 2 to 100 , inclusive , are not divisible by even integers greater than 1 ?
Options: a ) 5 , b ) 6 , c ) 8 , d ) 10 , e ) 50
Rationale: "the no will be of the form 2 ^ n to achieve this . since any other form will have odd no in the prime factorization . hence we need to find solution of n for equation - 1 < 2 ^ n < 101 . 2 ^ 6 = 64 , 2 ^ 7 - 128 hence n can take values from 1 to 6 . hence ans - ( b ) 6"
Answer: b
Problem: how many terminating zeroes does 200 ! have ?
Options: a ) 40 , b ) 48 , c ) 49 , d ) 55 , e ) 64
Rationale: we must realize that the trailing zeros depend on the number of 10 s ( i . e . 10 s or its multiples ) in the factorial . 10 = 5 * 2 , so we are looking for 5 s and 2 s . by virtue of being the more frequent number , we do n ' t need to count the number of 2 s , because there will always be equal or more 2 s than 5 s in a factorial . so , essentially we are looking for 5 s and its exponents . start with finding 5 s . 200 / 5 = 40 . so we have 40 fives ( and 40 twos ) which means we ' ll have 40 zeros . if the answer was a decimal , such as 201 / 5 , round down . now look for two 5 s , that is 25 . so , 200 / 25 = 8 , so we have eight more zeros . lastly , 5 ^ 3 = 125 , and we have one such number in 200 ! , which gives us one more trailing zero . therefore , we have 40 + 8 + 1 = 49 trailing zeros . answer : c
Answer: c
Problem: two airplanes fly the same exact route from atlanta to chicago . plane a flies 200 mph while plane b flies 300 mph ( assume that simple model versus actual acceleration / deceleration near the airports ) . if plane a took off exactly 40 minutes before plane b , after how many minutes will plane b overtake plane a ?
Options: a ) 65 , b ) 80 , c ) 90 , d ) 115 , e ) 120
Rationale: we can use some form of the equation d = rt [ distance = rate * time ] rate of plane a : 200 mph rate of plane b : 300 mph plane a will be in the sky 40 minutes longer than plane b ( 2 / 3 of an hour ) time of plane a when b overtakes it : t + ( 2 / 3 ) use 2 / 3 of an hour since the rate is in hours time of plane b when b overtakes a : t at the time that b overtakes a , they will have travelled the same distance so rt is equal for each plane : 200 * ( t + 2 / 3 ) = 300 * t 200 t + 400 / 3 = 300 t 400 / 3 = 100 t 4 / 3 = t this is 1 and a third hour , so 60 minutes + 20 minutes = 80 minutes b
Answer: b
Problem: a certain manufacturer produces items for which the production costs consist of annual fixed costs totaling $ 130,000 and variables costs averaging $ 11 per item . if the manufacturer ’ s selling price per item is $ 15 , how many items the manufacturer produce and sell to earn an annual profit of $ 150,000 ?
Options: a ) 2,858 , b ) 8,667 , c ) 21,429 , d ) 35,000 , e ) 70,000
Rationale: "let the items manufactured or sold bex 130000 + 11 x = 15 x - 150000 4 x = 280000 x = 70000 ans : e"
Answer: e
Problem: the area of a parallelogram is 450 sq m and its altitude is twice the corresponding base . then the length of the base is ?
Options: a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 15
Rationale: "2 x * x = 450 = > x = 15 answer : e"
Answer: e
Problem: a math teacher has 32 cards , each of which is in the shape of a geometric figure . half of the cards are rectangles , and a third of the cards are rhombuses . if 8 cards are squares , what is the maximum possible number of cards that re circles .
Options: a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 15
Rationale: "a square is a special kind of rhombus ( sides are perpendicular ) a square is a special kind of rectangles ( sides with same length ) among the 32 cards with have : 15 rectangles 10 rhombus 8 squares among the 15 rectangles , there could be 8 special ones ( with sides of same length ) that are squares . that lets at least 7 rectangles that are not square . among the 10 rectangles , there could be 8 special ones ( with sides perpendicular ) that are squares . that lets at least 2 rhombus that are not square . we have 8 squares . so the minimum different cards that represent a square , a rhombus or a rectangle is 2 + 7 + 8 = 17 which means that the maximum number of circles that you could have is 32 - 17 = 15 answer ( e )"
Answer: e
Problem: in one hour , a boat goes 15 km along the stream and 5 km against the stream . the speed of the boat in still water ( in km / hr ) is :
Options: a ) 2 , b ) 4 , c ) 10 , d ) 12 , e ) 15
Rationale: "sol . speed in still water = 1 / 2 ( 15 + 5 ) kmph = 10 kmph . answer c"
Answer: c
Problem: find the least number which when divided by 20 , 2535 and 40 leaves remainders 14 , 1929 and 34 respectively .
Options: a ) 1394 , b ) 1294 , c ) 1194 , d ) 1094 , e ) none of them
Rationale: here , ( 20 - 14 ) = 6 , ( 25 – 19 ) = 6 , ( 35 - 29 ) = 6 and ( 40 - 34 ) = 6 . therefore , required number = ( l . c . m . of 2025 , 3540 ) – 6 = 1394 . answer is a .
Answer: a
Problem: bag a contains red , white and blue marbles such that the red to white marble ratio is 1 : 3 and the white to blue marble ratio is 2 : 3 . bag b contains red and white marbles in the ratio of 1 : 4 . together , the two bags contain 28 white marbles . how many red marbles could be in bag a ?
Options: a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 7
Rationale: "# of red marbles in bag a can be either 2 or 6 . no 2 in the choices , so 6 . d . bag a : r : w : b = 2 : 6 : 9 bag b r : w = 1 : 4 6 x + 4 y = 28 i . e 3 x + 2 y = 14 x has to be odd to make an odd sum from the eq . x = 2 , y = 4 or x = 4 , y = 1 so r can be 2 x i . e 4 or 8 . answer c"
Answer: c
Problem: when average age of 22 members are 0 , how many members greater than 0 ?
Options: a ) 17 , b ) 21 , c ) 22 , d ) 24 , e ) 25
Rationale: average of 22 numbers = 0 . sum of 22 numbers ( 0 x 22 ) = 0 . it is quite possible that 21 of these numbers may be positive and if their sum is a then 22 nd number is ( - a ) answer is 21 ( b )
Answer: b
Problem: if 10 spiders make 5 webs in 5 days , then how many days are needed for 1 spider to make 1 web ?
Options: a ) 10 , b ) 9 , c ) 5 , d ) 15 , e ) 20
Rationale: "explanation : let , 1 spider make 1 web in x days . more spiders , less days ( indirect proportion ) more webs , more days ( direct proportion ) hence we can write as ( spiders ) 10 : 1 ( webs ) 1 : 5 } : : x : 5 â ‡ ’ 10 ã — 1 ã — 5 = 1 ã — 5 ã — x â ‡ ’ x = 10 answer : option a"
Answer: a
Problem: in a certain egg - processing plant , every egg must be inspected , and is either accepted for processing or rejected . for every 96 eggs accepted for processing , 4 eggs are rejected . if , on a particular day , 12 additional eggs were accepted , but the overall number of eggs inspected remained the same , the ratio of those accepted to those rejected would be 99 to 1 . how many e eggs does the plant process per day ?
Options: a ) 100 , b ) 300 , c ) 400 , d ) 3,000 , e ) 4,000
Rationale: "straight pluggin in for me . as usual , i started with c and got the answer . lets ' back calculate and see what we get let us consider eggs processed each day to be 400 so initial ratio of eggs processed and rejected is 96 : 4 or 24 : 1 so out of 400 eggs , there will be 384 eggs processed and 16 rejected . now if the no . of eggs inspected remain and 12 more eggs get accepted that means there e = 384 + 12 = 396 eggs accepted and 4 rejected . . . and the ratio will be 99 : 1 bingo . . . this is what the questions says . . . . its always a good idea to start with c ."
Answer: c
Problem: ram and shakil run a race of 2000 m . first ram gives shakil a start of 200 m and beats him by 1 minute . next ram gives shakil a start of 6 min and is beaten by 1000 meters . find the time in minutes in which ram and shakil can run the race seperately .
Options: a ) 12.18 , b ) 10.18 , c ) 10.12 , d ) 8.1 , e ) 8.2
Rationale: let r be velocity of ram and s be the velocity of shakil then ( 2000 / r ) = ( 1800 / s ) - 60 . . . . . . ( 1 ) second case distance shakil travel in 6 min = 360 s again equating time for both 1000 / r = ( 2000 - 360 s ) / s . . . . . . . ( 2 ) solving ( 1 ) & ( 2 ) we get s = 20 / 6 putting value in first we get time = 480 sec = 8 min answer : d
Answer: d
Problem: ravi and kavi start a business by investing â ‚ ¹ 4000 and â ‚ ¹ 72000 , respectively . find the ratio of their profits at the end of year .
Options: a ) 2 : 28 , b ) 5 : 28 , c ) 7 : 18 , d ) 1 : 18 , e ) 3 : 18
Rationale: "ratio of profit = ratio of investments = 4000 : 72000 = 1 : 18 answer : d"
Answer: d
Problem: 108 . triangle a ’ s base is 12 % greater than the base of triangle b , and a ’ s height is 12 % less than the height of triangle b . the area of triangle a is what percent less or more than the area of triangle b ?
Options: a ) 9 % less , b ) 0.7 % less , c ) equal to each other , d ) 1 % more , e ) 9 % more
Rationale: "wish the question specified that we are talking about corresponding height . base of a = 13 / 12 * base of b height of a = 11 / 12 * height of b area of a = ( 1 / 2 ) * base of a * height of a = 13 / 12 * 11 / 12 * area of b = 143 / 144 * area of b area of a is 0.7 % less than the area of b . answer ( b )"
Answer: b
Problem: two airplanes take off from one airfield at noon . one flies due east at 204 miles per hour while the other flies directly northeast at 283 miles per hour . approximately how many miles apart are the airplanes at 2 p . m . ?
Options: a ) 166 , b ) 408 , c ) 400 , d ) 483 , e ) 566
Rationale: b in two hours : the plane flying east will be 408 miles away from airport . the other plane will be 566 miles away from airport . 566 / 408 = ~ 1.4 = ~ sqrt ( 2 ) this means that planes formed a right isocheles triangle = > sides of such triangles relate as 1 : 1 : sqrt ( 2 ) = > the planes are 408 miles apart . b
Answer: b
Problem: the triplicate ratio of 2 : 5 is ?
Options: a ) 1 : 0 , b ) 8 : 125 , c ) 1 : 7 , d ) 1 : 2 , e ) 1 : 1
Rationale: "23 : 53 = 8 : 125 answer : b"
Answer: b
Problem: how many two - digit numbers that do not contain the digits 1 or 4 are there ?
Options: a ) 112 , b ) 56 , c ) 64 , d ) 49 , e ) 84
Rationale: "we can have 7 digits ( 2 , 3,5 , 6,7 , 8,9 ) for the first place ( tenth ' s place ) . and similarly 8 digits for the unit digit . ( 0,2 , 3,5 , 6,7 , 8,9 ) so in total 7 * 8 = 56 hence b"
Answer: b
Problem: marla starts running around a circular track at the same time nick starts walking around the same circular track . marla completes 10 laps around the track per hour and nick completes 5 laps around the track per hour . how many minutes after marla and nick begin moving will marla have completed 4 more laps around the track than nick ?
Options: a ) 5 , b ) 48 , c ) 12 , d ) 15 , e ) 20
Rationale: maria ' s rate - 10 laps per hour - - > 10 / 60 laps / min nick ' s rate - 5 laps per hour - - > 5 / 60 laps / min lets set equations : 10 / 60 * t = 4 ( since maria had to run 4 laps before nick would start ) 5 / 60 * t = 0 ( hick has just started and has n ' t run any lap yet ) ( 10 / 60 - 5 / 60 ) * t = 4 - 0 ( since nick was chasing maria ) t = 48 min needed maria to run 4 laps answer : b
Answer: b
Problem: the average expenditure of a labourer for 6 months was 75 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income i
Options: a ) 70 , b ) 72 , c ) 75 , d ) 78 , e ) 80
Rationale: "income of 6 months = ( 6 × 75 ) – debt = 450 – debt income of the man for next 4 months = 4 × 60 + debt + 30 = 270 + debt ∴ income of 10 months = 720 average monthly income = 720 ÷ 10 = 72 answer b"
Answer: b
Problem: dick and jane each saved $ 2,500 in 1989 . in 1990 dick saved 9 percent more than in 1989 , and together he and jane saved a total of $ 5200 . approximately what percent less did jane save in 1990 than in 1989 ?
Options: a ) 1 % , b ) 4 % , c ) 6 % , d ) 8 % , e ) 9 %
Rationale: "1990 dick saved = $ 2725 jane saved = $ 2475 ( jane saved $ 25 less than she did the prior year ) jane saved approximately $ 25 / $ 2500 ( 1 % ) less in 1990 answer : a"
Answer: a
Problem: seed mixture x is 40 % ryegrass and 60 % bluegrass by weight ; seed mixture y is 25 % ryegrass and 75 % fescue . if a mixture of x and y contains 31 % ryegrass , what percent of the weight of the mixture is from mixture x ?
Options: a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 60 %
Rationale: "31 % is 6 % - points above 25 % and 9 % - points below 40 % . thus the ratio of mixture y to mixture x is 3 : 2 . the percent of mixture x is 2 / 5 = 40 % . the answer is c ."
Answer: c
Problem: in a tree , 3 / 7 of the birds are parrots while the rest are toucans . if 2 / 3 of the parrots are female and 7 / 10 of the toucans are female , what fraction of the birds in the tree are male ?
Options: a ) 3 / 7 , b ) 11 / 35 , c ) 16 / 35 , d ) 31 / 70 , e ) 37 / 70
Rationale: "let x be the number of birds in the tree . the fraction of birds that are male parrots is ( 1 / 3 ) ( 3 / 7 ) = 1 / 7 . the fraction of birds that are male toucans is ( 3 / 10 ) ( 4 / 7 ) = 6 / 35 . the total fraction of male birds is 1 / 7 + 6 / 35 = 11 / 35 . the answer is b ."
Answer: b
Problem: income and expenditure of a person are in the ratio 10 : 8 . if the income of the person is rs . 10000 , then find his savings ?
Options: a ) 2003 , b ) 2002 , c ) 2000 , d ) 2001 , e ) 2004
Rationale: "let the income and the expenditure of the person be rs . 10 x and rs . 8 x respectively . income , 10 x = 10000 = > x = 1000 savings = income - expenditure = 10 x - 8 x = 2 x = 2 ( 1000 ) so , savings = rs . 2000 . answer : c"
Answer: c
Problem: in a school with 600 students , the average age of the boys is 12 years and that of the girls is 11 years . if the average age of the school is 11 years 9 months , then the number of girls in the school is
Options: a ) 150 , b ) 200 , c ) 250 , d ) 350 , e ) none
Rationale: "sol . let the number of grils be x . then , number of boys = ( 600 - x ) . then , ( 11 3 / 4 × 600 ) ⇔ 11 x + 12 ( 600 - x ) ⇔ x = 7200 - 7050 ⇔ 150 . answer a"
Answer: a
Problem: when sold at a 50 % discount , a sweater nets the merchant a 40 % profit on the wholesale cost at which he initially purchased the item . by what % is the sweater marked up from wholesale at its normal retail price ?
Options: a ) 20 % , b ) 64.29 % , c ) 70 % , d ) 80 % , e ) 100 %
Rationale: we should be careful about what are we measuring % on / what is the base . . let the marked up price = 100 . . selling price = 100 - 50 % of 100 = 50 . . profit = 40 % . . therefore the wholesale purchase cost = x . . . . 1.4 x = 50 or x = 35.71 . . . marked price was 100 so . . . so answer is 64.29 % . . b
Answer: b
Problem: one fourth of a solution that was 12 % sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight . the second solution was what percent sugar by weight ?
Options: a ) 34 % , b ) 28 % , c ) 22 % , d ) 18 % , e ) 8.5 %
Rationale: "instead of using complex calculations and remembering formulae , why dont u directly get to weighted average . 3 parts of 12 % + 1 part of x ( unknown ) % = 4 parts of 16 % = > x % = 64 % - 36 % = 28 % ans b it is ."
Answer: b
Problem: a palindrome is a number that reads the same forward and backward , such as 144 . how many odd , 4 - digit numbers are palindromes ?
Options: a ) 40 , b ) 45 , c ) 50 , d ) 62 , e ) 2500
Rationale: "a palindrome is a number that reads the same forward and backward . examples of four digit palindromes are 1221 , 4334 , 2222 etc you basically get to choose the first two digits and you repeat them in opposite order . say , you choose 45 as your first two digits . the next two digits are 54 and the number is 4554 . also , you need only odd palindromes . this means that you need an odd digit at the end i . e . 1 / 3 / 5 / 7 / 9 . this means that you need to start the number with an odd digit . only then will it end with an odd digit . in how many ways can you pick two digits such that the first one is an odd digit ? the first digit can be selected in 5 ways . ( 1 / 3 / 5 / 7 / 9 ) the second digit can be selected in 10 ways . ( 0 / 1 / 2 / 3 . . . 8 / 9 ) total = 5 * 12 = 62 ways d"
Answer: d