Cheukting/mathqa huggingface dataset:
Question (general): the sum of 3 consecutive odd natural numbers , each divisible by 3 is 72 . what is the largest among them ?
Options: a ) 21 , b ) 24 , c ) 27 , d ) 36 , e ) 45
Rationale: let the 3 consecutive odd numbers each divisible by 3 be x , x + 3 , x + 6 given , x + x + 3 + x + 6 = 72 3 x = 63 x = 21 the numbers are 21 , 24 , 27 therefore the largest number is 27 answer c .
Answer: c
Question (general): if sum of two no . is 12 and subtraction to that is 4 . so how much min . should added to smaller no . to make it even ?
Options: a ) 1 , b ) 2 , c ) 0.5 , d ) 1.5 , e ) 0
Rationale: let smaller no . be x . bigger no . = 12 - x . 12 - x - x = 4 . 12 - 2 x = 4 2 x = 8 , x = 4 . so 4 is itself even no . so , zero should be added . answer e
Answer: e
Question (gain): consider a lady took a loan from a bank at the rate of 12 % p . a . simple interest . after 3 years she had to pay rs . 5400 interest only for the period . the principal amount borrowed by her was
Options: a ) rs . 2000 , b ) rs . 10000 , c ) rs . 15000 , d ) rs . 20000 , e ) none of these
Rationale: "explanation : principal = rs . ( 100 × 5400 / 12 × 3 ) = > rs . 15,000 . answer : c"
Answer: c
Question (general): a store reported total sales of $ 385 million for february of this year . if the total sales for the same month last year was $ 320 million , approximately what was the percent increase q in sales ?
Options: a ) 2 % , b ) 17 % , c ) 20 % , d ) 65 % , e ) 83 %
Rationale: "last year ' s sales = $ 320 million ; this year ' s sales = $ 385 million ; increase q = $ 65 million . now , 20 % of $ 320 million is $ 64 million , which is very close to actual increase of $ 65 million . answer : c ."
Answer: c
Question (physics): one computer can upload 120 megabytes worth of data in 6 seconds . two computers , including this one , working together , can upload 1300 megabytes worth of data in 42 seconds . how long would it take for the second computer , working on its own , to upload 120 megabytes of data ?
Options: a ) 6 , b ) 7 , c ) 9.13 , d ) 11 , e ) 13
Rationale: "since the first computer can upload 120 megabytes worth of data in 6 seconds then in 6 * 7 = 42 seconds it can upload 7 * 120 = 840 megabytes worth of data , hence the second compute in 42 seconds uploads 1300 - 840 = 460 megabytes worth of data . the second computer can upload 120 megabytes of data in 9.13 seconds . answer : c ."
Answer: c
Question (general): set k contains every multiple of 6 from 18 to 306 inclusive . if w is the median of set k , and x is the average ( arithmetic mean ) of set k , what is the value of w - x ?
Options: a ) - 6 , b ) - 3 , c ) 0 , d ) 3 , e ) 6
Rationale: set k = { 18 , 24,30 , . . . . 306 } let the no . of terms be n . so we know the ap formulae to calculate the the tn term : tn = a + ( n - 1 ) * d where d = common difference of the terms . 306 = 18 + ( n - 1 ) * 6 n = 49 . so the set k consist of n terms . the median of the set having odd nos of elements is ( n + 1 ) / 2 . , which in this case is 25 . let ' s find the 25 th term using the same formulae again : t 25 = 18 + ( 25 - 1 ) * 6 t 25 = 162 so , the median of the set k is 162 i . e . w = 162 . now . lets find the average ( arithmetic mean ) of the set . for that we need to find the sum of all the elements first , lets call it s . since , set k is nothing but a arithmetic progression series having first element ( a ) as 18 , common difference ( d ) as 6 and no . of terms ( n ) as 49 . using the formulae to calculate sum of an ap series , which is s = n / 2 [ 2 a + ( n - 1 ) * d ] , we will calculate the sum . so , s = 49 / 2 [ 2 * 18 + ( 49 - 1 ) * 6 ] this gives us s = 7938 . now arithmetic mean of set k = 7938 / no . of terms = 7938 / 49 = 162 . so x = 162 . now , ( w - x ) = ( 162 - 162 ) = 0 . therefore correct answer = c
Answer: c
Question (general): # p is defined as 2 p + 20 for any number p . what is p , if # ( # ( # p ) ) = 4 ?
Options: a ) – 108 , b ) – 44 , c ) 10 , d ) - 17 , e ) 18
Rationale: # p = 2 p + 20 - - - > # ( # p ) = 2 ( 2 p + 20 ) + 20 = 4 p + 60 and thus # ( 4 p + 60 ) = 2 ( 4 p + 60 ) + 20 = 8 p + 140 = 4 - - - > 8 p = - 136 - - - > p = - 17 , d is the correct answer .
Answer: d
Question (general): if n = 3 ^ 0.15 and n ^ b = 9 , b must equal
Options: a ) 3 / 80 , b ) 3 / 5 , c ) 4 , d ) 5 / 3 , e ) 40 / 3
Rationale: "15 / 100 = 3 / 20 n = 3 ^ 3 / 20 n ^ b = 3 ^ 2 ( 3 ^ 3 / 20 ) ^ b = 3 ^ 2 b = 40 / 3 answer : e"
Answer: e
Question (physics): two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 36 hours and 16 hours respectively . the ratio of their speeds is ?
Options: a ) 4 : 5 , b ) 4 : 3 , c ) 4 : 6 , d ) 4 : 9 , e ) 4 : 2
Rationale: "let us name the trains a and b . then , ( a ' s speed ) : ( b ' s speed ) = √ b : √ a = √ 16 : √ 36 = 4 : 6 answer : c"
Answer: c
Question (gain): john makes $ 55 a week from his job . he earns a raise andnow makes $ 60 a week . what is the % increase ?
Options: a ) 16 % , b ) 9.66 % , c ) 9.09 % , d ) 9.98 % , e ) 17 %
Rationale: "increase = ( 5 / 55 ) * 100 = ( 1 / 11 ) * 100 = 9.09 % . c"
Answer: c
Question (general): what is the sum of all the prime numbers greater than 10 but less than 40 ?
Options: a ) 190 , b ) 189 , c ) 180 , d ) 199 , e ) 1981
Rationale: "required sum = ( 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 ) = 180 note : 1 is not a prime number answer c"
Answer: c
Question (gain): the value of a machine depreciates at the rate of 10 % every year . it was purchased 3 years ago . if its present value is rs . 8748 , its purchase price was
Options: a ) rs . 12008 , b ) rs . 12000 , c ) rs . 12002 , d ) rs . 1229 , e ) rs . 12021
Rationale: explanation : = rs . 12000 answer : b
Answer: b
Question (general): find 95 × × 97
Options: a ) 92 / 198 , b ) 92 / 12 , c ) 92 / 13 , d ) 92 / 10 , e ) 92 / 15
Rationale: "here both numbers are less than 100 . so they are deficient of - 5 and - 3 compared with 100 . so answer : e"
Answer: e
Question (general): what is the units digit of 17 ^ 73 × 13 ^ 82 × 11 ^ 87 ?
Options: a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8
Rationale: "to find : the units digit of 17 ^ 83 × 13 ^ 82 × 11 ^ 87 let ' s reduce the clutter and simplify the product ( 7 ^ 83 ) ( 3 ^ 82 ) ( 1 ^ 87 ) 7 has a cyclicity of 4 : the last digit of any positive power of 7 repeats itself after every 4 th power so 7 ^ 5 has the same last digit as 7 ^ 1 , 7 ^ 9 , 7 ^ 13 thus , 7 ^ 83 has the same last digit as 7 ^ 3 , 7 ^ 7 , 7 ^ 11 i . e . 3 3 has a cyclicity of 4 : exactly the same routine as above thus , 3 ^ 82 has the same last digit as 3 ^ 2 , 3 ^ 6 , 3 ^ 10 i . e . 9 any power of 1 will result in 1 as the last digit so , product of our last digits = 3 x 9 x 1 = 27 . . . . last digit is 8 correct option : e"
Answer: e
Question (general): the price of commodity x increases by 30 cents every year , while the price of commodity y increases by 20 cents every year . if in 2001 , the price of commodity x was $ 4.20 and the price of commodity y was $ 4.40 , in which year will commodity x cost 70 cents more than the commodity y ?
Options: a ) 2010 , b ) 2011 , c ) 2012 , d ) 2013 , e ) 2014
Rationale: "the cost of commodity x increases by 10 cents per year relative to commodity y . the price of x must gain 20 + 70 = 90 cents on commodity y , which will take 9 years . the answer is a ."
Answer: a
Question (physics): how many bricks , each measuring 100 cm x 11.25 cm x 6 cm , will be needed to build a wall of 8 m x 6 m x 22.5 cm ?
Options: a ) 1600 , b ) 2400 , c ) 5500 , d ) 7400 , e ) 3400
Rationale: "number of bricks = volume of the wall / volume of 1 brick = ( 800 x 600 x 22.5 ) / ( 100 x 11.25 x 6 ) = 1600 answer : a"
Answer: a
Question (general): real - estate salesman z is selling a house at a 20 percent discount from its retail price . real - estate salesman x vows to match this price , and then offers an additional 10 percent discount . real - estate salesman y decides to average the prices of salesmen z and x , then offer an additional 25 percent discount . salesman y ' s final price is what fraction of salesman x ' s final price ?
Options: a ) 72 / 80 , b ) 72 / 100 , c ) 57 / 80 , d ) 19 / 24 , e ) 76 / 100
Rationale: "let retail price = $ 100 . z sells with 20 % discount . so z sells at $ 80 . x matches the price and gives additional 10 % . so , x sells at 80 - ( 10 % of 80 ) = 80 - 8 = $ 72 y avgs x and z [ ( 72 + 80 ) / 2 = 76 ] . y gives additional 25 % discount . so , y sells at 76 - ( 25 % of 76 ) = 76 - 19 = $ 57 now , ratio of y to x = 57 / 72 = 19 / 24 answer will be d ."
Answer: d
Question (general): if a - b = 3 and a ^ 2 + b ^ 2 = 28 , find the value of ab .
Options: a ) 19 , b ) 15 , c ) 10 , d ) 8 , e ) 13
Rationale: "2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 25 - 9 = 16 ab = 8 . answer is d ."
Answer: d
Question (physics): a group of men decided to do a work in 20 days , but 10 of them became absent . if the rest of the group did the work in 40 days , find the original number of men ?
Options: a ) 20 , b ) 50 , c ) 40 , d ) 100 , e ) 25
Rationale: "original number of men = 10 * 40 / ( 40 - 20 ) = 20 answer is a"
Answer: a
Question (gain): mr . karan borrowed a certain amount at 6 % per annum simple interest for 9 years . after 9 years , he returned rs . 8310 / - . find out the amount that he borrowed .
Options: a ) s . 5266 , b ) s . 5396 , c ) s . 5228 , d ) s . 5218 , e ) s . 52192
Rationale: "explanation : let us assume mr . karan borrowed amount is rs . a . ( the principal ) by formula of simple interest , s . i . = prt / 100 where p = the principal , r = rate of interest as a % , t = time in years s . i . = ( p * 6 * 9 ) / 100 = 54 p / 100 amount = principal + s . i . 8310 = p + ( 54 p / 100 ) 8310 = ( 100 p + 54 p ) / 100 8310 = 154 p / 100 p = ( 8310 * 100 ) / 154 = rs . 5396.104 answer : b"
Answer: b
Question (gain): an article is bought for rs . 980 and sold for rs . 788 , find the loss percent ?
Options: a ) 16 % , b ) 19 % , c ) 17 % , d ) 13 % , e ) 12 %
Rationale: "980 - - - - 192 100 - - - - ? = > 19 % answer : b"
Answer: b
Question (probability): out of 15 students in a class , 7 are from maharashtra , 5 are from karnataka , and 3 are from goa . four students are to be selected at random . what are the chances that at least one is from karnataka ?
Options: a ) 12 / 13 , b ) 11 / 13 , c ) 10 / 15 , d ) 1 / 15 , e ) 4 / 14
Rationale: total possible ways of selecting 4 students out of 15 = 15 c 4 = ( 15 * 14 * 13 * 12 ) / ( 4 * 3 * 2 * 1 ) = 1365 the number of ways of selecting a 4 students in which no student belongs to karnataka = 10 c 4 number of ways of selecting atleast one student from karnataka = 15 c 4 - 10 c 4 = 1155 probability = 1155 / 1365 = 11 / 13 answer is b
Answer: b
Question (general): suppose that f ( x ) is a function such that for every real number x , i ) f ( x ) + f ( 1 - x ) = 10 and ( ii ) f ( 1 + x ) = 3 + f ( x ) . then f ( x ) + f ( - x ) must equal
Options: a ) 7 , b ) 9 , c ) 10 , d ) 11 , e ) 12
Rationale: since ( ii ) holds for every real number x , it will remain valid if we replace x with - x . therefore , f ( 1 - x ) = 3 + f ( - x ) . now , from ( i ) , 10 = f ( x ) + f ( 1 - x ) = f ( x ) + 3 + f ( - x ) so that f ( x ) + f ( - x ) = 10 - 3 = 7 . ( observe that f ( x ) = 3 x + 4 satisfies the conditions in the problem . ) correct answer a
Answer: a
Question (general): each child has 12 crayons and 24 apples . if there are 18 children , how many crayons are there in total ?
Options: a ) 220 , b ) 65 , c ) 216 , d ) 219 , e ) 230
Rationale: 12 * 18 = 216 . answer is c .
Answer: c
Question (general): in a certain warehouse , 90 percent of the packages weigh less than 75 pounds , and a total of 48 packages weigh less than 25 pounds . if 80 percent of the packages weigh at least 25 pounds , how many of the packages weigh at least 25 pounds but less than 75 pounds ?
Options: a ) 8 , b ) 64 , c ) 168 , d ) 102 , e ) 144
Rationale: "if 80 % of the packages weigh at least 25 pounds this means that 20 % of the packages weigh less than 25 pounds let t = total number of packages so , 20 % of t = # of packages that weigh less than 25 pounds 48 packages weigh less than 25 pounds great . so , 20 % of t = 48 rewrite to get : 0.2 t = 48 solve : t = 240 90 % of the packages weigh less than 75 pounds so , 90 % oft = number of packages that weigh less than 75 pounds 90 % of 240 = 216 , so 216 packages weigh less than 75 pounds of those 216 packages that weigh less than 75 pounds , 48 packages weigh less than 25 pounds . so , the number of packages that weight between 25 and 75 pounds = 216 - 48 = 168 = c"
Answer: c
Question (general): what is the units digit of the expression 14 ^ 7 − 19 ^ 4 ?
Options: a ) 0 , b ) 3 , c ) 4 , d ) 6 , e ) 8
Rationale: i think answer on this one should be e too . since we know that 14 ^ 7 > 19 ^ 4 , as will said one should always check if the number is positive .
Answer: e
Question (geometry): there are two concentric circles such that the area of the outer circle is four times the area of the inner circle . let a , b and c be three distinct points on the perimeter of the outer circle such that ab and ac are tangents to the inner circle . if the area of the outer circle is 12 square centimetres then the area ( in square centimetres ) of the triangle abc would be
Options: a ) π √ ​ 12
b ) 9 / π
c ) 9 √ 3 / π
d ) 6 √ 3 / π
e ) none of these
Rationale: explanation : outer circle : since π r ² = 12 , r = √ ( 12 / π ) = 2 √ ( 3 / π ) . inner circle : since the area is 1 / 4 the area of the outer circle , π r ² = 3 and r = √ ( 3 / π ) ao is the radius of the outer circle and od is the radius of the inner circle . a line tangent to a circle and a radius drawn to the point of tangency form a right angle . thus , ∆ aod is a right triangle . hypotenuse ao is twice the length of leg od . a right triangle whose hypotenuse is twice the length of a leg is a 30 - 60 - 90 triangle . thus , ∆ aod is a 30 - 60 - 90 triangle . in a 30 - 60 - 90 triangle , the sides are proportioned x : x √ 3 : 2 x . thus , ad = √ 3 * √ ( 3 / π ) * = 3 / √ π . using the same reasoning for ∆ ocd , we can deduce that cd = 3 / √ π . thus , ac = 2 * 3 / √ π = 6 / √ π . the formula for the area of an equilateral triangle = ( s ² √ 3 ) / 4 . since ac = 6 / √ π , the area of ∆ abc = ( 6 / √ π ) ² * √ 3 / 4 = 36 / π * √ 3 / 4 = ( 9 √ 3 ) / π . answer : c
Answer: c
Question (general): if 425 / 999 = 0.125 , what is the 81 st digit to the right of the decimal point in the decimal equivalent of the fraction ?
Options: a ) 0 , b ) 2 , c ) 3 , d ) 4 , e ) 5
Rationale: 0 . [ u ] 125 [ / u = 0.425425425 . . . . . every 3 rd digit is 5 and every multiple of 3 will be the digit 5 . since 81 is multiple of 3 , the 81 st digit is 5 . answer e
Answer: e
Question (gain): a and b start a business , with a investing the total capital of rs . 500000 , on the condition that b pays a interest @ 10 % per annum on his half of the capital . a is a working partner and receives rs . 15000 per month from the total profit and any profit remaining is equally shared by both of them . at the end of the year , it was found that the income of a is twice that of b . find the total profit for the year ?
Options: a ) a ) 590280 , b ) b ) 590100 , c ) c ) 591900 , d ) d ) 590000 , e ) e ) 927000
Rationale: "interest received by a from b = 10 % of half of rs . 500000 = 10 % * 250000 = 25000 . amount received by a per annum for being a working partner = 15000 * 12 = rs . 180000 . let ' p ' be the part of the remaining profit that a receives as his share . total income of a = ( 25000 + 180000 + p ) total income of b = only his share from the remaining profit = ' p ' , as a and b share the remaining profit equally . income of a = twice the income of b ( 25000 + 180000 + p ) = 2 ( p ) p = 205000 total profit = 2 p + 180000 = 2 * 205000 + 180000 = 590000 answer : d"
Answer: d
Question (general): if p and q are prime numbers , how many divisors does the product p ^ 2 * q ^ 5 have ?
Options: a ) 8 , b ) 10 , c ) 12 , d ) 15 , e ) 18
Rationale: "when a number n = a ^ x * b ^ y , where a and b are prime numbers , and x , y are positive integers , the number of divisors of n = ( x + 1 ) ( y + 1 ) therefore , the answer is e . 3 * 6 = 18"
Answer: e
Question (physics): the figure above shows the dimensions of a semicircular cross section of a one - way tunnel . the single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel . if vehicles must clear the top of the tunnel by at least ½ foot when they are inside the traffic lane , what should be the limit l on the height of vehicles that are allowed to use the tunnel ?
Options: a ) 5 ½ ft , b ) 7 ½ ft , c ) 8 ½ ft , d ) 9 ½ ft , e ) 10 ft
Rationale: "let ' s label the midpoint of the circle o . since the base of the semi - circle is 20 , we know that the diameter is 20 and , accordingly , the radius is 10 . we also know that the traffic lane is 12 feet long and there ' s an equal amount of space on either side , so the traffic lane extends 6 feet on either side of o . let ' s call the leftmost point on the base of the traffic lane a . so , the distance oa is 6 . now draw a line straight up from a to the top of the tunnel . let ' s label the point at which the line intersects the circle b . the answer to the question will , therefore , be the height ab - . 5 feet ( we need to leave . 5 feet of clearance ) . here ' s the key to solving the question : if we draw a line from o to b , that line is a radius of the circle and , therefore , has length 10 . we now have right triangle oab ( the right angle is at point a ) , with leg oa = 6 and hypotenuse ob = 10 . we can now solve for leg ab = 8 ( either by applying the pythagorean theorum or by applying the 3 / 4 / 5 special right triangle ratio ) . finally : ab = 8 , so the correct answer l is 8 - . 5 = 7.5 . . . choose ( b ) !"
Answer: b
Question (gain): a factory has 15 machines of equal efficiency in its factory . the annual manufacturing costs are rs . 42000 & establishment charges are rs . 12000 . the annual output of the company is rs . 70000 . the annual output and manufacturing costs are directly proportional to the number of machines . the shareholders get 12 . 512.5 profit , which is directly proportional to the annual output of the company . if 7 . 147.14 machines remain closed throughout the year , then the % decrease in the amountof profit of the shareholders would be ?
Options: a ) 12.5 % , b ) 13.5 % , c ) 14 % , d ) 15 % , e ) 16 %
Rationale: option ( b ) is correct original profit = 70,000 − 42,000 − 12,000 = 16 , 00070,000 − 42,000 − 12,000 = 16,000 if 7.14 % 7.14 % of 14 i . e . one of the machines closed throughout the year , then change in profit will be : = 1314 × [ 70,000 − 42,000 ] = 1314 × [ 70,000 − 42,000 ] = 14,000 = 14,000 thus , decrease in the profit % % = 200016000 × 100 = 200016000 × 100 = 12.5 % a
Answer: a
Question (other): the h . c . f of two numbers is 23 and the other two factors of their l . c . m are 11 and 12 . the larger of the two numbers is :
Options: a ) 338 , b ) 276 , c ) 322 , d ) 231 , e ) 121
Rationale: "clearly , the numbers are ( 23 * 11 ) and ( 23 * 12 ) . larger number = ( 23 * 12 ) = 276 . answer : b"
Answer: b
Question (gain): the present population of a town is 3600 . population increase rate is 20 % p . a . find the population of town before 2 years ?
Options: a ) 2500 , b ) 2100 , c ) 3500 , d ) 3600 , e ) 2050
Rationale: "p = 3600 r = 20 % required population of town = p / ( 1 + r / 100 ) ^ t = 3600 / ( 1 + 20 / 100 ) ^ 2 = 3600 / ( 6 / 5 ) ^ 2 = 2500 ( approximately ) answer is a"
Answer: a
Question (general): evaluate : 20 - 12 ÷ 4 × 2 =
Options: a ) a ) 12 , b ) b ) 24 , c ) c ) 36 , d ) d ) 48 , e ) e ) 60
Rationale: "according to order of operations , 12 ÷ 4 × 2 ( division and multiplication ) is done first from left to right 12 ÷ 4 × 2 = 3 × 2 = 6 hence 20 - 12 ÷ 4 × 2 = 20 - 6 = 14 correct answer is b ) 22"
Answer: b
Question (general): the average of a batsman for 40 innings is 50 runs . his highest score exceeds his lowest score by 172 runs . if these two innings are excluded , his average drops by 2 runs . find his highest score .
Options: a ) 172 , b ) 173 , c ) 174 , d ) 175 , e ) none of the above
Rationale: total runs = 40 × 50 = 2000 let his highest score be = x then his lowest score = x – 172 now 200 − x − ( x − 172 ) / 38 = 48 ⇒ 2 x = 2172 – 1824 ⇒ x = 174 answer c
Answer: c
Question (gain): a corporation paid $ 9 million in federal taxes on its first $ 50 million of gross profits and then $ 30 million in federal taxes on the next $ 150 million in gross profits . by approximately what percent did the ratio of federal taxes to gross profits increase from the first $ 50 million in profits to the next $ 150 million in profits ?
Options: a ) 6 % , b ) 14 % , c ) 20 % , d ) 23 % , e ) 14 %
Rationale: "difference in ratios = ( 30 / 150 ) - ( 9 / 50 ) = ( 1 / 50 ) % change = ( change ( 1 / 50 ) / original ratio ( 7 / 50 ) ) * 100 = 14 % answer - e"
Answer: e
Question (physics): the height of the wall is 6 times its width and length of the wall is 7 times its height . if the volume of the wall be 129024 cu . m . its width is
Options: a ) 4 m , b ) 5 m , c ) 6 m , d ) 7 m , e ) 8 m
Rationale: "explanation : let width = x then , height = 6 x and length = 42 x 42 x ã — 6 x ã — x = 129024 x = 8 answer : e"
Answer: e
Question (general): evaluate : 136.09 + 43.9
Options: a ) 1000 , b ) 3000 , c ) 7000 , d ) 5000 , e ) 2000
Rationale: "25 / 0.0005 = ( 25 * 10000 ) / ( 0.0005 * 10000 ) = 25000 / 5 = 5000 answer is d"
Answer: d
Question (other): 16 balls are numbered 1 to 16 . a ball is drawn and then another ball is drawn without replacement . what is the probability that both balls have even numbers ?
Options: a ) 4 / 11 , b ) 5 / 17 , c ) 6 / 25 , d ) 7 / 30 , e ) 8 / 37
Rationale: "p ( 1 st ball is even ) = 8 / 16 p ( 2 nd ball is also even ) = 7 / 15 p ( both balls are even ) = 8 / 16 * 7 / 15 = 7 / 30 the answer is d ."
Answer: d
Question (general): the number of years of service of the 8 employees in a production department are 15 , 10 , 9 , 17 , 6 , 3 , 14 and 16 . what is the range in the number of years of service of the 8 employees ?
Options: a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14
Rationale: = 17 - 3 = 14 answer e
Answer: e
Question (other): how many factors of 440 are odd numbers greater than 1 ?
Options: a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 7
Rationale: when factorized , 440 has 4 prime factors . of these prime factors 2 are odd and 2 are even . hence total number of odd factors is 2 * 2 ( 4 ) , which includes 2 . the total number of odd factors greater than 1 are 2 . ( option a )
Answer: a
Question (physics): annika hikes at a constant rate of 10 minutes per kilometer . she has hiked 2.5 kilometers east from the start of a hiking trail when she realizes that she has to be back at the start of the trail in 35 minutes . if annika continues east , then turns around and retraces her path to reach the start of the trail in exactly 35 minutes , for how many kilometers total did she hike east ?
Options: a ) 3.625 , b ) 3.5 , c ) 3 , d ) 4 , e ) 4.5
Rationale: set up two r x t = d cases . 1 . 1 / 10 km / min x t = 2.5 from which t = 25 mins . we know total journey time now is 35 + 25 = 60 the rate is the same ie 1 / 10 km / min . set up second r x t = d case . 1 / 10 km / min x 60 = 6 km now the total journey would be halved as distance would be same in each direction . 6 / 2 = 3 c
Answer: c