regisss/math_qa huggingface dataset:
Problem: in what time will a train 150 metres long cross an electric pole , if its speed be 122 km / hr ?
Options: a ) 2.42 , b ) 6.42 , c ) 5.42 , d ) 4.42 , e ) 3.42
Rationale: "solution speed = ( 122 x 5 / 18 ) m / sec = 33.88 m / sec time taken = ( 150 / 33.88 ) sec = 4.42 sec . answer d"
Answer: d
Problem: what is the sum of the greatest common factor and the lowest common multiple of 16 and 36 ?
Options: a ) 148 , b ) 222 , c ) 144 , d ) 157 , e ) 164
Rationale: "prime factorization of the given numbers 16 = 2 ^ 4 36 = 2 ^ 2 * 3 * 2 greatest common factor = 2 ^ 2 = 4 lowest common multiple = 2 ^ 4 * 3 ^ 2 = 144 sum = 4 + 144 = 148 answer a"
Answer: a
Problem: x is a positive integer less than 100 . when x is divided by 7 , the remainder is 1 ; when x is divided by 3 , the remainder is 2 . how many x are there ?
Options: a ) 14 , b ) 22 , c ) 23 , d ) 24 , e ) 25
Rationale: "the nubmer which when divided by 7 leaves remainder 1 should be of the form 7 k + 1 this number when divided by 3 leaves remainder 2 . so , ( 7 k + 1 ) - 2 should be divisible by 3 or 7 k - 1 should be divisible by 3 . we now put the values of k starting from 0 to find first number divisible by 3 we find 1 st number at k = 1 thus smallest number will be 7 ( 1 ) + 1 = 8 now , next number will be = 8 + lcm of 37 i . e 29 now we will find number of all such values less than 500 by using the formula for last term of an a . p 8 + ( n - 1 ) 21 = 100 n = 14.42 or n = 14 answer : - a"
Answer: a
Problem: a student has to obtain 33 % of the total marks to pass . he got 125 marks and failed by 40 marks . the maximum mark is
Options: a ) 600 , b ) 500 , c ) 400 , d ) 300 , e ) 800
Rationale: "the student got 125 marks and still failed by 40 marks = 125 + 40 = 165 therefore 33 % of the total marks = 165 100 % of the total = 165 * 100 / 33 = 55 max mark = 500 answer b"
Answer: b
Problem: the number of boxes in a warehouse can be divided evenly into 9 equal shipments by boat or 24 equal shipments by truck . what is the smallest number of boxes that could be in the warehouse ?
Options: a ) 27 , b ) 33 , c ) 72 , d ) 81 , e ) 162
Rationale: "answer is the lcm of 9 and 24 = 72 answer c"
Answer: c
Problem: an outlet pipe empties a tank which is full in 5 hours . if the inlet pipe is kept open , which lets water in at the rate of 8 litres / min then outlet pipe would take 3 hours longer . find the capacity of the tank .
Options: a ) 8600 litres , b ) 200 litres , c ) 12800 litres , d ) 6400 litres , e ) 13200 litres
Rationale: "let the rate of outlet pipe be x liters / hour ; rate of inlet pipe is 8 litres / min , or 8 * 60 = 480 liters / hour ; net outflow rate when both pipes operate would be x - 480 liters / hour . capacity of the tank = x * 5 hours = ( x - 480 ) * ( 5 + 3 ) hours 5 x = ( x - 480 ) * 8 - - > x = 1280 - - > capacity = 5 x = 6400 liters . answer : d"
Answer: d
Problem: a can do a piece of work in 3 hours ; b and c together can do it in 2 hours , which a and b together can do it in 2 hours . how long will c alone take to do it ?
Options: a ) 2 hours , b ) 5 hours , c ) 6 hours , d ) 4 hours , e ) 3 hours
Rationale: "a ' s 1 hour work = 1 / 3 ; ( b + c ) ' s 1 hour work = 1 / 2 ; ( a + b ) ' s 1 hour work = 1 / 2 ( a + b + c ) ' s 1 hour work = ( 1 / 3 + 1 / 2 ) = 5 / 6 c ' s 1 hour work = ( 5 / 6 - 1 / 2 ) = 1 / 3 c alone will take 3 hours to do the work . answer : e"
Answer: e
Problem: tony alone can paint a wall in 7 days and his friend roy alone can paint the same wall in 9 days . in how many days they can paint the wall working together ? round off the answer to the nearest integer .
Options: a ) 2 , b ) 4 , c ) 6 , d ) 7 , e ) 9
Rationale: sol . use formula ( xy / x + y ) so nearest value for 3.93 = 4 answer : option b
Answer: b
Problem: when the number 9 y 30012 is exactly divisible by 11 , what is the smallest whole number that can replace y ?
Options: a ) 1 , b ) 2 , c ) 5 , d ) 6 , e ) 8
Rationale: "the given number = 9 y 30012 sum of the odd places = 2 + 0 + 3 + 9 = 14 sum of the even places = 1 + 0 + y ( sum of the odd places ) - ( sum of even places ) = number ( exactly divisible by 11 ) 14 - ( 1 + y ) = divisible by 11 13 � y = divisible by 11 . y must be 2 , to make given number divisible by 11 . b"
Answer: b
Problem: a worker ' s daily wage is increased by 40 % and the new wage is $ 35 per day . what was the worker ' s daily wage before the increase ?
Options: a ) $ 20 , b ) $ 21 , c ) $ 24 , d ) $ 25 , e ) $ 28
Rationale: "let x be the daily wage before the increase . 1.4 x = $ 35 x = $ 25 the answer is d ."
Answer: d
Problem: for the past n days , the average ( arithmetic mean ) daily production at a company was 60 units . if today ' s production of 90 units raises the average to 65 units per day , what is the value of n ?
Options: a ) 30 , b ) 18 , c ) 10 , d ) 9 , e ) 5
Rationale: "( average production for n days ) * n = ( total production for n days ) - - > 60 n = ( total production for n days ) ; ( total production for n days ) + 90 = ( average production for n + 1 days ) * ( n + 1 ) - - > 60 n + 90 = 65 * ( n + 1 ) - - > n = 5 . or as 30 extra units increased the average for n + 1 days by 5 units per day then 30 / ( n + 1 ) = 5 - - > n = 5 . answer : e ."
Answer: e
Problem: a 7 - digit combination lock on a safe has zero exactly 3 times , does not have the digit 1 at all . what is the probability that exactly 3 of its digits are odd ?
Options: a ) 9 / 16 , b ) 1 / 2 , c ) 1 / 3 , d ) 1 / 4 , e ) 1 / 6
Rationale: remaining number of digits after three ' 0 ' digits = 4 constraint - no more ' 0 ' and no more ' 1 ' well , these four digits can be odd or even 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 16 ways to choose 3 odd digits out of total 4 digits = 4 c 3 = 4 4 * 1 / 16 = 1 / 4 d is the answer hi ! since we are talking about remaining 4 digits . each digit can either be odd or even ( total ways to select = 2 . desired outcome = 1 ) so let ' s say we want to choose first 3 digits as odd digits and last digit as even . the probability will be : - 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 16 but out of remaining 4 digit , any 3 can be odd and one even as there are no constraints in the question . therefore , total ways of choosing 3 odd digits out of 4 total digits = 4 c 3 so , the probability = 4 * 1 / 16 = 1 / 4 answer is d
Answer: d
Problem: area of a square is 1 / 2 hectare . the diagonal of the square is ?
Options: a ) 250 meter
b ) 100 meter
c ) 50 √ 2 meter
d ) 50 meter
e ) 35 meter
Rationale: area = 1 / 2 hectare = 10000 / 2 m 2 = 5000 m 2 again area = 1 / 2 x ( diagonal ) 2 so 1 / 2 x ( diagonal ) 2 = 5000 m 2 diagonal 2 = 10000 diagonal = 100 answer : b
Answer: b
Problem: the ratio of the cost price and the selling price is 4 : 5 . the profit percent is :
Options: a ) 67 % , b ) 45 % , c ) 25 % , d ) 70 % , e ) 50 %
Rationale: "c c . p . = rs . 4 x . then , s . p . = rs . 5 x gain = ( 5 x - 4 x ) = rs . x gain % = ( x * 100 ) / 4 x = 25 % ."
Answer: c
Problem: pipe a and pipe b fill water into a tank of capacity 5000 litres , at a rate of 200 l / min and 50 l / min . pipe c drains at a rate of 25 l / min . pipe a is open for 1 min and closed , then pipe b is open for 2 min and closed . further the pipe c is opened and drained for another 2 min . this process is repeated until the tank is filled . how long will it take to fill the tank ?
Options: a ) 120 , b ) 100 , c ) 140 , d ) 80 , e ) 90
Rationale: tank capacity : 5000 l , 1 st - 200 l / min for 1 min , volume filled : 200 l 2 nd - 100 l / min for 2 min , volume filled : 100 l 3 rd ( water draining ) : 25 l / min * 2 : 50 l total : ( 200 + 100 ) - 50 = 250 l filled for 1 cycle number of 250 in 5000 l tank : 1000 / 250 = 20 time taken to fill : 20 * total time = 20 * 5 = 100 ( option b )
Answer: b
Problem: a can do a piece of work in 10 days and b in 20 days . they began the work together but 5 days before the completion of the work , a leaves . the work was completed in ?
Options: a ) 15 days , b ) 10 days , c ) 12 days , d ) 19 days , e ) 13 days
Rationale: b ( x â € “ 5 ) / 10 + x / 20 = 1 x = 10 days
Answer: b
Problem: a car gets 20 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 130 kilometers ?
Options: a ) 8.5 gallons , b ) 7.5 gallons , c ) 6.5 gallons , d ) 5.5 gallons , e ) 4.5 gallons
Rationale: "each 20 kilometers , 1 gallon is needed . we need to know how many 20 kilometers are there in 130 kilometers ? 130 / 20 = 6.5 * 1 gallon = 6.5 gallons correct answer c"
Answer: c
Problem: 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 )
Options: a ) 40 m 2 , b ) 44 m 2 , c ) 52.8 m 2 , d ) 36 m 2 , e ) none of these
Rationale: "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"
Answer: c
Problem: at what price must an article costing rs . 200 be marked in order that after deducting 10 % from the list price . it may be sold at a profit of 20 % on the cost price ?
Options: a ) 62.5 , b ) 62.0 , c ) 266.67 , d ) 62.1 , e ) 62.2
Rationale: "cp = 200 sp = 200 * ( 120 / 100 ) = 240 mp * ( 90 / 100 ) = 240 mp = 266.67 answer : c"
Answer: c
Problem: evaluate : 11 + sqrt ( - 4 + 6 × 4 ÷ 3 ) = ?
Options: a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14
Rationale: "according to order of operations , inner brackets first where 6 × 4 ÷ 3 is first calculated since it has a multiplication and a division . 6 × 4 ÷ 3 = 24 ÷ 3 = 8 hence 11 + sqrt ( - 4 + 6 × 4 ÷ 3 ) = 11 + sqrt ( - 4 + 8 ) = 11 + sqrt ( 4 ) = 11 + 2 = 13 correct answer d ) 13"
Answer: d
Problem: the lcm and hcf of two numbers are 1000 and 100 respectively . find the larger of the two numbers if their sum is 1100 .
Options: a ) 1000 , b ) 10 , c ) 567 , d ) 57 , e ) 678
Rationale: "there are 2 approaches in solving this . methode 1 . hcf * lcm = the actual number . 1000 * 100 = 100000 so the answer which we are looking for has to be a factor of 100000 . so among the options shortlist the answers by eliminating those numbers which is not divisible by 100000 . and then take the highest number as the answer as the question asks abt the highest number . answer is a"
Answer: a
Problem: ramesh purchased a refrigerator for rs . 12500 after getting a discount of 20 % on the labelled price . he spent rs . 125 on transport and rs . 250 on installation . at what price should it be sold so that the profit earned would be 15 % if no discount was offered ?
Options: a ) 17608 , b ) 17606 , c ) 17604 , d ) 18400 , e ) 117601
Rationale: "price at which the tv set is bought = rs . 12,500 discount offered = 20 % marked price = 12500 * 100 / 80 = rs . 15625 the total amount spent on transport and installation = 125 + 250 = rs . 375 \ total price of tv set = 15625 + 375 = rs . 16000 the price at which the tv should be sold to get a profit of 15 % if no discount was offered = 16000 * 115 / 100 = rs . 18400 . answer : d"
Answer: d
Problem: in a school 10 % of the boys are same in number as 1 / 6 th of the girls . what is the ratio of boys to the girls in the school ?
Options: a ) 5 : 3 , b ) 2 : 3 , c ) 1 : 4 , d ) 3 : 7 , e ) 2 : 5
Rationale: "10 % of b = 1 / 6 g 10 b / 100 = g / 6 b = 5 g / 3 b / g = 5 / 3 b : g = 5 : 3 answer is a"
Answer: a
Problem: the number of positive integer solutions for the equation x + y + z + t = 15 is
Options: a ) 654 , b ) 584 , c ) 384 , d ) 364 , e ) 394
Rationale: the number of positive integer solutions for the equatio fx 1 + x 2 + ⋯ + xn = k ( k - 1 ) c ( n - 1 ) - where k is the number and n is number of variable in the equation . 15 - 1 c 4 - 1 = 14 c 3 = 364 answer : d
Answer: d
Problem: in a certain pond , 50 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 ` s the approximate number of fish in the pond ?
Options: a ) 400 , b ) 625 , c ) 1250 , d ) 2500 , e ) 10 000
Rationale: "if x is total number of fish in the pond : 4 = 50 / x * 100 = > x = 1250 so answer is c"
Answer: c
Problem: if the cost price of 12 pencils is equal to the selling price of 8 pencils , the gain percent is :
Options: a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70
Rationale: c 50 % let c . p . of each pencil be $ 1 . then , c . p . of 8 pencils = $ 8 ; s . p . of 8 pencils = $ 12 . gain % = 4 / 8 * 100 = 50 %
Answer: c
Problem: a and b together can do a work in 9 days . if a alone can do it in 18 days . in how many days can b alone do it ?
Options: a ) 10 , b ) 99 , c ) 18 , d ) 55 , e ) 21
Rationale: "1 / 9 – 1 / 18 = 1 / 18 = > 18 answer : c"
Answer: c
Problem: solve below question 6 x - 1 = - 13
Options: a ) - 8 , b ) - 2 , c ) - 5 , d ) - 4 , e ) 1
Rationale: 1 . subtract 1 from both sides : 6 x - 1 + 1 = - 13 + 1 2 . simplify both sides : 6 x = - 12 3 . divide both sides by 6 : 4 . simplify both sides : x = - 2 b
Answer: b
Problem: if n = ( 32 ) ^ 43 + ( 43 ) ^ 33 what is the units digit of n ?
Options: a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8
Rationale: "first of all , the units digit of ( 32 ) ^ 43 is the same as that of 3 ^ 43 and the units digit of ( 43 ) ^ 33 is the same as that of 3 ^ 33 . so , we need to find the units digit of 3 ^ 43 + 3 ^ 33 . next , the units digit of 3 in positive integer power repeats in blocks of four { 3 , 9 , 7 , 1 } : 3 ^ 1 = 3 ( the units digit is 3 ) 3 ^ 2 = 9 ( the units digit is 9 ) 3 ^ 3 = 27 ( the units digit is 7 ) 3 ^ 4 = 81 ( the units digit is 1 ) 3 ^ 5 = 243 ( the units digit is 3 again ! ) . . . thus : the units digit of 3 ^ 43 is the same as the units digit of 3 ^ 3 , so 7 ( 43 divided by the cyclicity of 4 gives the remainder of 3 ) . the units digit of 3 ^ 33 is the same as the units digit of 3 ^ 1 , so 3 ( 33 divided by the cyclicity of 4 gives the remainder of 1 ) . therefore the units digit of ( 23 ) ^ 43 + ( 43 ) ^ 33 is 5 + 3 = 0 . answer : e ."
Answer: e
Problem: if 125 ( 5 ^ y ) = 1 then y =
Options: a ) – 2 , b ) – 1 , c ) 0 , d ) 1 , e ) - 3
Rationale: 5 ^ x = 1 / 125 5 ^ x = 1 / 5 ^ 3 5 ^ x = 5 ^ - 3 x = - 3 e
Answer: e
Problem: a certain store sells all maps at one price and all books at another price . on monday the store sold 12 maps and 8 books for a total of $ 38.00 , and on tuesday the store sold 20 maps and 12 books for a total of $ 60.00 . at this store , how much less does a map sell for than a book ?
Options: a ) $ 0.25 , b ) $ 0.50 , c ) $ 0.75 , d ) $ 1.00 , e ) $ l . 25
Rationale: "12 x + 8 y = 38 20 x + 12 y = 60 multiply eq 1 by 5 and eq 2 by 3 60 x + 40 y = 38 * 5 60 x + 36 y = 60 * 3 subtracting 1 from 2 4 y = 190 - 180 y = 2.5 therefore x = 1.5 difference in price = 1 d"
Answer: d
Problem: 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
Options: a ) rs . 7000 , b ) rs . 8000 , c ) rs . 8500 , d ) rs . 9000 , e ) none
Rationale: "answer average salary = 8000 + 5000 + 11000 + 7000 + 9000 / 5 = rs . 8000 correct option : b"
Answer: b
Problem: a car averages 75 mph for the first 4 hours of a trip and averages 60 mph for each additional hour . the average speed for the entire trip was 70 mph . how many hours long is the trip ?
Options: a ) 6 , b ) 78 , c ) 8 , d ) 9 , e ) 10
Rationale: "let the time for which car averages 60 mph = t 70 * ( t + 4 ) = 75 * 4 + 60 t = > 10 t = 20 = > t = 2 total duration of the trip = 4 + 2 = 6 answer a"
Answer: a
Problem: linda spent 3 / 4 of her savings on furniture and the rest on a tv . if the tv cost her $ 210 , what were her original savings ?
Options: a ) $ 840 , b ) $ 800 , c ) $ 1000 , d ) $ 700 , e ) $ 1500
Rationale: "if linda spent 3 / 4 of her savings on furnitute , the rest 4 / 4 - 3 / 4 = 1 / 4 on a tv but the tv cost her $ 210 . so 1 / 4 of her savings is $ 210 . so her original savings are 4 times $ 210 = $ 840 correct answer a"
Answer: a
Problem: the duplicate ratio of 1 : 4 is ?
Options: a ) 1 : 2 , b ) 1 : 4 , c ) 1 : 8 , d ) 1 : 18 , e ) 1 : 13
Rationale: "1 ^ 2 : 4 ^ 2 = 2 : 16 = 1 : 8 answer : c"
Answer: c
Problem: find the cost of fencing around a circular field of diameter 20 m at the rate of rs . 1.50 a meter ?
Options: a ) 287 , b ) 94.2 , c ) 156 , d ) 158 , e ) 267
Rationale: "2 * 22 / 7 * 10 = 62.8 62.8 * 1 1 / 2 = rs . 94.2 answer : b"
Answer: b
Problem: a 60 kg metal bar made of alloy of tin and silver lost 6 kg of its weight in the water . 10 kg of tin loses 1.375 kg in the water ; 5 kg of silver loses 0.375 kg . what is the ratio of tin to silver in the bar ?
Options: a ) 1 / 4 , b ) 2 / 5 , c ) 1 / 2 , d ) 3 / 5 , e ) 2 / 3
Rationale: "you can simply use this formula to avoid confusion : w 1 / w 2 = ( a 2 - aavg ) / ( avg - a 1 ) here is how you will find the values of a 1 an a 2 . we have an overall loss ( average loss ) . the average loss is 6 kg when 60 kg alloy is immersed . this is a loss of ( 6 / 60 ) * 100 = 10 % . this is aavg the loss of tin is 1.375 kg for every 10 kg . this means it loses ( 1.375 / 10 ) * 100 = 13.75 % of its weight in water . this is a 1 . the loss of silver is . 375 kg for every 5 kg . this means it loses ( . 375 / 5 ) * 100 = 7.5 % of its weight in water . this is a 2 . weight of tin / weight of silver = ( silver ' s loss - avg loss ) / ( avg loss - tin ' s loss ) x / y = ( 7.5 - 10 ) / ( 10 - 13.75 ) = 2 / 3 e"
Answer: e
Problem: a walks at 10 kmph and 10 hours after his start , b cycles after him at 20 kmph . how far from the start does b catch up with a ?
Options: a ) 100 km , b ) 150 km , c ) 50 km , d ) 120 km , e ) 200 km
Rationale: "suppose after x km from the start b catches up with a . then , the difference in the time taken by a to cover x km and that taken by b to cover x km is 10 hours . x / 10 - x / 20 = 10 x = 200 km answer is e"
Answer: e
Problem: the sum of the non - prime numbers between 60 and 70 , non - inclusive , is
Options: a ) 387 , b ) 429 , c ) 457 , d ) 499 , e ) 536
Rationale: "sum of consecutive integers from 61 to 69 , inclusive = = = = > ( a 1 + an ) / 2 * # of terms = ( 61 + 69 ) / 2 * 9 = 65 * 9 = 585 sum of non - prime numbers b / w 60 and 70 , non inclusive = = = > 585 - 128 ( i . e . , 61 + 67 , being the prime # s in the range ) = 457 answer : c"
Answer: c
Problem: how many seconds will a 500 meter long train moving with a speed of 63 km / hr , take to cross a man walking with a speed of 3 km / hr in the direction of the train ?
Options: a ) 15 s , b ) 20 s , c ) 25 s , d ) 30 s , e ) 35 s
Rationale: distance = 500 m speed = 63 - 3 km / hr = 60 km / hr = 600 / 36 m / s = 50 / 3 m / s time taken = distance / speed = 500 / ( 50 / 3 ) = 30 s answer is d .
Answer: d
Problem: evaluate 50 % of 450 + 45 % of 500
Options: a ) 232 , b ) 450 , c ) 252 , d ) 262 , e ) 272
Rationale: "explanation : = ( 50 / 100 ) * 450 + ( 45 / 100 ) * 500 = 450 option b"
Answer: b
Problem: ifaequals the sum of the even integers from 2 to 20 , inclusive , andbequals the sum of the odd integers from 1 to 19 , inclusive , what is the value of b - a ?
Options: a ) - 10 , b ) 10 , c ) - 19 , d ) 20 , e ) 21
Rationale: "yes ! there is really a faster way to solve it . sum of consecutive odd or even integers = ( no . of odd or even ints ) * ( first int + last int ) / 2 here a = sum of even ints from 2 to 20 , inclusive number of even ints = 10 , first int + last int = 2 + 20 = 22 a = 10 * 22 / 2 = 110 b = sum of odd ints from 1 to 19 , inclusive number of odd ints = 10 , first int + last int = 1 + 19 = 20 a = 10 * 20 / 2 = 100 b - a = 100 - 110 = - 10 ans : a"
Answer: a
Problem: the angle of elevation of a ladder leaning against a wall is 60 º and the foot of the ladder is 4.6 m away from the wall . the length of the ladder is :
Options: a ) 2.3 m , b ) 4.6 m , c ) 7.8 m , d ) 9.2 m , e ) 10.1
Rationale: "let ab be the wall and bc be the ladder . then , acb = 60 º and ac = 4.6 m . ac / bc = cos 60 º = 1 / 2 bc = 2 x ac = ( 2 x 4.6 ) m = 9.2 m . answer : option d"
Answer: d
Problem: what is the cost of leveling the field in the form of parallelogram at the rate of rs . 50 / 10 sq . metre , whose base & perpendicular distance from the other side being 94 m & 54 m respectively ?
Options: a ) s . 6420 , b ) s . 3563 , c ) s . 2400 , d ) s . 5480 , e ) s . 5076
Rationale: "area of the parallelogram = length of the base * perpendicular height = 94 * 54 = 5076 m . total cost of levelling = rs . 5076 e"
Answer: e
Problem: what is the minimum value of | x - 4 | + | x + 8 | + | x - 5 | ?
Options: a ) - 13 , b ) 13 , c ) 5 , d ) 7 , e ) 8
Rationale: "a can not be the answer as all the three terms are in modulus and hence the answer will be non negative . | x - 4 | > = 0 - - > minimum occurs at x = 4 | x + 8 | > = 0 - - > minimum occurs at x = - 8 | x - 5 | > = 0 - - > minimum occurs at x = 5 x = - 8 - - > result = 12 + 0 + 13 = 25 . also any negative value will push the combined value of | x - 4 | + | x - 5 | to a value > 9 . x = 4 - - > result = 0 + 12 + 1 = 13 x = 5 - - > result = 1 + 13 + 0 = 14 x = 8 - - > result = 4 + 16 + 3 = 23 so minimum value of the expression occurs at x = 4 and the resultant value = 13 answer : b"
Answer: b
Problem: divide rs . 1600 among a , b and c so that a receives 1 / 4 as much as b and c together and b receives 3 / 4 as a and c together . a ' s share is ?
Options: a ) 350 , b ) 375 , c ) 320 , d ) 530 , e ) 230
Rationale: "a + b + c = 1600 a = 1 / 4 ( b + c ) ; b = 3 / 4 ( a + c ) a / ( b + c ) = 1 / 4 a = 1 / 5 * 1600 = > 320 answer : c"
Answer: c
Problem: the population of a town is 10000 . it increases annually at the rate of 15 % p . a . what will be its population after 2 years ?
Options: a ) 13300 , b ) 13350 , c ) 13225 , d ) 13500 , e ) 13600
Rationale: "formula : ( after = 100 denominator ago = 100 numerator ) 10000 ã — 115 / 100 ã — 115 / 100 = 13225 c"
Answer: c
Problem: a man ' s speed with the current is 16 km / hr and the speed of the current is 3.2 km / hr . the man ' s speed against the current is
Options: a ) 9 , b ) 9.6 , c ) 11 , d ) 12 , e ) 13
Rationale: "man ' s rate in still water = ( 16 - 3.2 ) km / hr = 12.8 km / hr . man ' s rate against the current = ( 12.8 - 3.2 ) km / hr = 9.6 km / hr . answer : b"
Answer: b
Problem: a card game called “ high - low ” divides a deck of 52 playing cards into 2 types , “ high ” cards and “ low ” cards . there are an equal number of “ high ” cards and “ low ” cards in the deck and “ high ” cards are worth 2 points , while “ low ” cards are worth 1 point . if you draw cards one at a time , how many ways can you draw “ high ” and “ low ” cards to earn 6 points if you must draw exactly 3 “ low ” cards ?
Options: a ) 1 , b ) 20 , c ) 3 , d ) 4 , e ) 5
Rationale: "great question ravih . this is a permutations problem ( order matters ) with repeating elements . given thatlowcards are worth 1 pt andhigh cards 2 pts , and you must draw 3 low cards , we know that you must also draw 1 high card . the formula for permutations problems with repeating elements isn ! / a ! b ! . . . where n represents the number of elements in the group and a , b , etc . represent the number of times that repeating elements are repeated . here there are 4 elements and thelowcard is repeated 3 times . as a result , the formula is : 5 ! / 3 ! which represents ( 5 * 4 * 3 * 2 * 1 ) / ( 3 * 2 * 1 ) which simplifies to just 20 , giving you answer b ."
Answer: b
Problem: a card is drawn from a pack of 52 cards . the probability of not getting a face card ?
Options: a ) 9 / 13 , b ) 5 / 12 , c ) 3 / 13 , d ) 6 / 17 , e ) 1 / 13
Rationale: clearly in the 52 cards other than face cards = 52 - 16 = 36 probability of not getting a face card = 36 / 52 = 9 / 13 correct option is a
Answer: a
Problem: what number is 64 more than two - fourth of itself ?
Options: a ) 128 , b ) 225 , c ) 288 , d ) 324 , e ) 336
Rationale: "2 / 4 x + 64 = x that means 64 = 2 / 4 x x = 64 * 2 = 128 a is the answer"
Answer: a
Problem: in 10 years , a will be twice as old as b was 10 years ago . if a is now 6 years older than b the present age of b is
Options: a ) 37 , b ) 36 , c ) 39 , d ) 40 , e ) 41
Rationale: "let present age of a be a and b be b a + 10 = 2 * ( b - 10 ) = > 2 b - a = 30 . . . . . . ( i ) a = b + 6 = > 2 b - b - 6 = 30 b = 36 so the present age of b is 36 years answer : b"
Answer: b
Problem: a train running at the speed of 56 km / hr crosses a pole in 9 sec . what is the length of the train ?
Options: a ) 140 m , b ) 786 m , c ) 566 m , d ) 546 m , e ) 445 m
Rationale: "speed = 56 * 5 / 18 = 140 / 9 m / sec length of the train = speed * time = 140 / 9 * 9 = 140 m answer : a"
Answer: a
Problem: the cross - section of a cannel is a trapezium in shape . if the cannel is 11 m wide at the top and 9 m wide at the bottom and the area of cross - section is 960 sq m , the depth of cannel is ?
Options: a ) 69 , b ) 92 , c ) 94 , d ) 49 , e ) 96
Rationale: "1 / 2 * d ( 11 + 9 ) = 960 d = 96 answer : e"
Answer: e
Problem: suppose 6 monkeys take 6 minutes to eat 6 bananas . how many monkeys would it take to eat 18 bananas in 18 minutes ?
Options: a ) 9 , b ) 10 , c ) 11 , d ) 6 , e ) 13
Rationale: "one monkey takes 6 min to eat 1 banana , so in 18 mins 1 monkey will eat 3 bananas , so for 18 bananas in 18 min we need 18 / 3 = 6 monkeys answer : d"
Answer: d
Problem: 6 ) a marketing firm determined that , of 240 households surveyed , 80 used neither brand a nor brand b soap . 60 used only brand a soap and for every household that used both brands of soap , 3 used only brand b soap . how many of the 200 household surveyed used both brands of soap ?
Options: a ) a ) 15 , b ) b ) 20 , c ) c ) 25 , d ) d ) 40 , e ) e ) 45
Rationale: "160 = at least one of soap a or b both brands = x brand b = 3 x = > 60 + x + 3 x = 160 = > 4 x = 100 = > x = 25 answer - c"
Answer: c
Problem: if 20 % of a number is equal to three - fifth of another number , what is the ratio of first number to the second number ?
Options: a ) 3 : 1 , b ) 2 : 1 , c ) 1 : 3 , d ) 1 : 2 , e ) 2 : 3
Rationale: "let 20 % of a = 3 / 5 b . then , 20 a / 100 = 3 b / 5 = > 1 a / 5 = 3 b / 5 a / b = ( 3 / 5 * 5 / 1 ) = 3 / 1 a : b = 3 : 1 . answer : a"
Answer: a
Problem: the sum of the ages of 5 children born at the intervals of 3 years each is 80 years . what is the age of the youngest child ?
Options: a ) 2 years , b ) 4 years , c ) 6 years , d ) 8 years , e ) 10 years
Rationale: "let the ages of the children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 80 5 x = 50 = > x = 10 . age of youngest child = x = 10 years . answer : e"
Answer: e
Problem: a runs 1 2 ⁄ 3 times as fast as b . if a gives b a start of 80 m , how far must the winning post be so that a and b might reach it at the same time ?
Options: a ) 200 m , b ) 250 m , c ) 450 m , d ) 350 m , e ) 270 m
Rationale: speed of a : speed of b = 5 ⁄ 3 : 1 = 5 : 3 i . e . , in a race of 5 m , a runs 5 m and b runs 3 m i . e . , a gains 2 m over b in a race of 5 m = > a gains 80 m over b in a race of 52 × 80 = 200 m hence , winning point should be 200 m away from the starting point . answer : a
Answer: a
Problem: in a division , a student took 42 as divisor instead of 36 . his answer was 24 . the correct answer is -
Options: a ) 42 , b ) 32 , c ) 48 , d ) 28 , e ) 38
Rationale: "x / 42 = 24 . x = 24 * 42 . so correct answer would be , ( 24 * 42 ) / 36 = 38 . answer : e"
Answer: e
Problem: hyderabad chennai mumbai delhi bangalore trivandram ; hyderabad chennai mumbai delhi bangalore trivandram in the table above , what is the least number of table entries that are needed to show the mileage between each city and each of the other 5 cities ?
Options: a ) 20 , b ) 18 , c ) 14 , d ) 15 , e ) 16
Rationale: easy way to go about this problem is we have 6 * 6 = 36 enteries in table the least number of enteries would be ( 36 - 6 ) / 2 since 6 enteries represent the distances between same points . alternatively this can be solved as combination problem . correct answer d
Answer: d
Problem: the speed of a boat in still water in 20 km / hr and the rate of current is 5 km / hr . the distance travelled downstream in 24 minutes is :
Options: a ) 10.6 , b ) 10.2 , c ) 10.8 , d ) 10.4 , e ) 10.0
Rationale: "speed downstream = ( 20 + 5 ) = 25 kmph time = 24 minutes = 24 / 60 hour = 2 / 5 hour distance travelled = time × speed = 2 / 5 × 25 = 10.0 km answer is e ."
Answer: e
Problem: the average weight of 8 person ' s increases by 3.5 kg when a new person comes in place of one of them weighing 65 kg . what is the weight of the new person ?
Options: a ) 93 kg , b ) 50 kg , c ) 85 kg , d ) 80 kg , e ) 60 kg
Rationale: "explanation : total increase in weight = 8 ã — 3.5 = 28 if x is the weight of the new person , total increase in weight = x â ˆ ’ 65 = > 28 = x - 65 = > x = 28 + 65 = 93 answer : option a"
Answer: a
Problem: a river 5 m deep and 19 m wide is flowing at the rate of 4 kmph the amount of water that runs into the sea per minute is ?
Options: a ) 4500 , b ) 6333 , c ) 6576 , d ) 6345 , e ) 2882
Rationale: "( 4000 * 5 * 19 ) / 60 = 6333 m 3 answer : b"
Answer: b
Problem: if the number is divided by 3 , it reduced by 24 . the number is
Options: a ) a ) 51 , b ) b ) 50 , c ) c ) 45 , d ) d ) 40 , e ) e ) 36
Rationale: "explanation : let the number be x . then , x - ( x / 3 ) = 24 = > 2 x / 3 = 24 = > x = 36 answer : option e"
Answer: e
Problem: 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 ?
Options: a ) $ 21 , b ) $ 27 , c ) $ 31 , d ) $ 43 , e ) it can not be determined
Rationale: "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"
Answer: d
Problem: the prices of tea and coffee per kg were the same in june . in july the price of coffee shot up by 20 % and that of tea dropped by 20 % . if in july , a mixture containing equal quantities of tea and coffee costs 40 / kg . how much did a kg of coffee cost in june ?
Options: a ) 40 , b ) 60 , c ) 80 , d ) 100 , e ) 120
Rationale: "let the price of tea and coffee be x per kg in june . price of tea in july = 1.2 x price of coffee in july = 0.8 x . in july the price of 1 / 2 kg ( 400 gm ) of tea and 1 / 2 kg ( 400 gm ) of coffee ( equal quantities ) = 40 1.2 x ( 1 / 2 ) + 0.8 x ( 1 / 2 ) = 40 = > x = 40 thus proved . . . option a ."
Answer: a
Problem: solve below question 4 x + 1 = 37
Options: a ) - 8 , b ) - 9 , c ) - 5 , d ) 9 , e ) 1
Rationale: 1 . subtract 1 from both sides : 4 x + 1 - 1 = 37 - 1 2 . simplify both sides : 4 x = 36 3 . divide both sides by 4 : 4 . simplify both sides : x = 9 d
Answer: d
Problem: thabo owns exactly 220 books , and each book is either paperback fiction , paperback nonfiction , or hardcover nonfiction . if he owns 20 more paperback nonfiction books than hardcover nonfiction books , and twice as many paperback fiction books as paperback nonfiction books , how many hardcover books nonfiction books does thabo own ?
Options: a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50
Rationale: "i think we can use double - matrix method and solve using only one variable . our goal is to find the number of hardcover nonfiction books . let that number be x . we are given that all 140 books are either paperback fiction , paperback nonfiction , or hardcover nonfiction . this implies that number of hardcover fiction books is 0 . double - matrix : p = paperback ; h = hardcover ; f = fiction ; nf = nonfiction p h total f 2 x + 40 0 nf x + 20 x total 3 x + 60 x 220 3 x + 60 + x = 220 x = 40 answer ( d . )"
Answer: d
Problem: 10 men do a work in 10 days . how many men are needed to finish the work in 5 days ?
Options: a ) 50 , b ) 20 , c ) 30 , d ) 10 , e ) 15
Rationale: "men required to finish the work in 5 days = 10 * 10 / 5 = 20 answer is b"
Answer: b
Problem: the average age of a family of 6 members is 28 years . if the age of the youngest member is 7 years , what was the average age of the family at the birth of the youngest member ?
Options: a ) 15 , b ) 18 , c ) 21 , d ) 12 , e ) 19
Rationale: "present age of total members = 6 x 28 = 168 7 yrs back their ages were = 6 x 7 = 42 ages at the birth of youngest member = 168 - 42 = 126 therefore , avg age at the birth of youngest member = 126 / 6 = 21 . answer : c"
Answer: c
Problem: there are 84 supermarkets in the fgh chain . all of them are either in the us or canada . if there are 14 more fgh supermarkets in the us than in canada , how many fgh supermarkets are there in the us ?
Options: a ) 20 , b ) 31 , c ) 42 , d ) 49 , e ) 64
Rationale: x + ( x - 14 ) = 84 - - > x = 49 . answer : d .
Answer: d
Problem: find out the square of a number which when doubled exceeds its one eighth by 15 ?
Options: a ) 16 , b ) 25 , c ) 19 , d ) 64 , e ) 17
Rationale: "let the number be p , then the square will be p ^ 2 according to question : 2 p = ( p / 8 ) + 15 = > 16 p = p + 120 = > p = 8 p ^ 2 = 8 ^ 2 = 64 . answer : d"
Answer: d
Problem: the h . c . f . of two numbers is 5 and their l . c . m . is 150 . if one of the numbers is 25 , then the other is :
Options: a ) 30 , b ) 28 , c ) 24 , d ) 20 , e ) 35
Rationale: "product of two nos = product of hcf and lcm let one number = x = > 25 * x = 5 * 150 x = 5 * 150 / 25 = 30 answer a"
Answer: a
Problem: a and b are two partially filled buckets of water . if 4 liters are transferred from a to b , then a would contain one - third of the amount of water in b . alternatively , if 5 liters are transferred from b to a , b would contain one - half of the amount of water in a . bucket a contains how many liters of water ?
Options: a ) 47 / 5 , b ) 13 , c ) 17 , d ) 21 , e ) 23
Rationale: "let a contains a , b contains b liters so , ( a - 4 ) / ( b + 4 ) = 1 / 3 . . . . . . . ( 1 ) again , ( b - 5 ) / ( a + 5 ) = 1 / 2 . . . . . . . . . . ( 2 ) from ( 1 ) ( 2 ) we find a = 47 / 5 ans : ( a )"
Answer: a
Problem: a sporting good store sells one type of baseball bat and one type of baseball . the cost for 2 bats and 4 balls is $ 220 . the cost for 1 bat and 6 balls is $ 190 , as well . if someone were to buy an equal number of bats and balls , at most how many bats can he purchase if he has a budget of $ 270 for the purchase ?
Options: a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
Rationale: "imo it should be c that is 4 reason : formed an equation . . . bat = b ball = c 2 b + 4 c = 220 1 b + 6 c = 190 solving both we get b that is bat = 70 and c that is ball = 20 new equation 270 to be divided in equal 3 b + 3 c = 270 3 * 70 + 3 * 20 = 270 210 + 60 = 270"
Answer: c
Problem: in a coconut grove , ( x + 2 ) trees yield 30 nuts per year , x trees yield 120 nuts per year and ( x – 2 ) trees yield 180 nuts per year . if the average yield per year per tree be 100 , find x .
Options: a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 10
Rationale: "( x + 2 ) × 30 + x × 120 + ( x − 2 ) × 180 / ( x + 2 ) + x + ( x − 2 ) = 100 ⇒ 330 x − 300 / 3 x = 100 ⇒ 30 x = 300 ⇒ x = 10 answer e"
Answer: e
Problem: a football player scores 2 goals in his fifth match thus increasing his average goals score by 0.3 . the total number of goals in his 5 matches would be
Options: a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12
Rationale: "while this question can be solved with a rather straight - forward algebra approach ( as the other posters have noted ) , it can also be solved by testing the answers . one of those numbers must be the total number of goals . . . from a tactical standpoint , it ' s best to test either answer b or answer d , so if the answer is not correct , then you would have a gauge for whether you should gohigherorlowerwith your next test . here , i ' ll start with answer a = 4 goals if . . . . total goals = 4 goals 5 th game = 2 goals 1 st 4 games = 2 goals avg . for 1 st 4 games = 2 / 4 = 0.5 goal / game avg . for all 5 games = 4 / 5 = 0.8 goals / game this is an exact match for what we ' re told in the prompt , so answer a must be the answer ."
Answer: a
Problem: a salt solution contains 15 grams of salt per 1000 cubic centimeters of solution . if 25 cubic centimeters of the solution were poured into an empty container , how many grams of salt would be in the container ?
Options: a ) 0.550
b ) 0.350
c ) 0.450
d ) 0.700
e ) 0.375
Rationale: we are given that a salt solution contains 15 grams of salt per 1000 cubic centimeters of solution . since we are dealing with a solution , we know that the grams of salt is proportional to the number of cubic centimeters of solution . thus , to determine how many grams of salt would be in the container when we have 25 cubic centimeters of solution , we can set up a proportion . we can say : “ 15 grams of salt is to 1000 cubic centimeters of solution as x grams of salt is to 25 cubic centimeters of solution . ” let ’ s now set up the proportion and solve for x . 15 / 1000 = x / 25 when we cross multiply we obtain : ( 15 ) ( 25 ) = 1000 x 375 = 1000 x 0.375 = x there are 0.375 grams of salt in the solution in the container . the answer is e .
Answer: e
Problem: jennifer had $ 120 to spend on herself . she spent 1 / 5 of the money on a sandwich , 1 / 6 for a ticket to a museum , and 1 / 2 of it on a book . how much money does jennifer have left over ?
Options: a ) $ 16 , b ) $ 14 , c ) $ 5 , d ) $ 15 , e ) $ 0
Rationale: "1 / 5 x $ 120 = $ 24 for sandwich 1 / 6 x $ 120 = $ 20 for museum 1 / 2 x $ 120 = $ 60 for book $ 24 + $ 20 + $ 60 = $ 104 spent $ 120 - $ 104 = $ 16 left over correct answer a"
Answer: a
Problem: if the sum of two numbers is 55 and the h . c . f . and l . c . m of these numbers are 5 and 120 respectively , then the sum of the reciprocals of the numbers is equal to :
Options: a ) 55 / 601 , b ) 601 / 55 , c ) 11 / 120 , d ) 120 / 11 , e ) 120 / 13
Rationale: "given a + b = 55 , hcm and lcm is 5 and 120 axb = lcmx hcf 600 = 120 x 5 1 / a + 1 / b = a + b / ab = > 55 / 600 = > 11 / 120 answer : c"
Answer: c
Problem: the banker ' s gain on a bill due due 1 year hence at 12 % per annum is rs . 7.8 . the true discount is
Options: a ) rs . 72 , b ) rs . 65 , c ) rs . 54 , d ) rs . 50 , e ) none
Rationale: "solution t . d = [ b . g x 100 / r x t ] = rs . ( 7.8 x 100 / 12 x 1 ) = rs . 65 . answer b"
Answer: b
Problem: a man engaged a servant on the condition that he would pay him rs . 900 and a uniform after 1 year service . he served only for 9 months and receiveduniform and rs . 650 , find the priceof the uniform ?
Options: a ) rs . 90 , b ) rs . 100 , c ) rs . 120 , d ) rs . 150 , e ) rs . 160
Rationale: "9 / 12 = 3 / 4 * 900 = 675 650 - - - - - - - - - - - - - 25 1 / 4 - - - - - - - - 25 1 - - - - - - - - - ? = > rs . 100 b"
Answer: b
Problem: 2 men and 4 boys can complete a work in 4 days . 5 men and 6 boys can complete the same work in 3 days . the work done by 2 boys is equal to the work of how many men ?
Options: a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8
Rationale: exp : ( 2 m + 4 b ) ’ s one day ’ s work = 1 / 4 ( 5 m + 6 b ) ’ s one day ’ s work = 1 / 3 = > ( 8 m + 16 b ) ’ s one day ’ s work = ( 15 m + 18 b ) ’ s one day ’ s work = > 7 men ’ s work = 2 boy ’ s work so , we should be employ 5 more men to be completed the work . answer : b
Answer: b
Problem: 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
Options: a ) rs . 55.50 , b ) rs . 67.50 , c ) rs . 86.50 , d ) rs . 101.50 , e ) none of these
Rationale: "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"
Answer: d
Problem: of 1000 surveyed students , 20 % of those who read book a also read book b and 25 % of those who read book b also read book a . if each student read at least one of the books , what is the difference between the number of students who read only book a and the number of students who read only book b ?
Options: a ) 120 , b ) 125 , c ) 130 , d ) 145 , e ) 155
Rationale: "say the number of students who read book a is a and the number of students who read book b is b . given that 20 % of those who read book a also read book b and 25 % of those who read book b also read book a , so the number of students who read both books is 0.2 a = 0.25 b - - > a = 1.25 b . since each student read at least one of the books then { total } = { a } + { b } - { both } - - > 1000 = 1.25 b + b - 0.25 b - - > b = 500 , a = 1.25 b = 625 and { both } = 0.25 b = 125 . the number of students who read only book a is { a } - { both } = 625 - 125 = 500 ; the number of students who read only book b is { b } - { both } = 500 - 125 = 375 the difference is 500 - 375 = 125 . answer : b ."
Answer: b
Problem: a and b finish the job in 15 days . while a , b and c can finish it in 3 days . c alone will finish the job in
Options: a ) 4.75 days , b ) 3.75 days , c ) 6.50 days , d ) 7.50 days , e ) 5.50 days
Rationale: "explanation : 3 = ( 15 * x ) / ( 15 + x ) 45 + 3 x = 15 x 12 x = 45 x = 3.75 answer : option b"
Answer: b
Problem: the sum of 3 integers a , b and c is 60 . a is one third of the sum of b and c and b is one fifth of the sum of a and c . what is c ?
Options: a ) 40 , b ) 45 , c ) 35 , d ) 50 , e ) 55
Rationale: a is one third of the sum of b and c . . or a = 1 / 3 ( b + c ) . . . but a + b + c = 60 so 1 / 3 ( b + c ) + b + c = 60 4 ( b + c ) / 3 = 60 b + c = 60 * 3 / 4 = 45 a = 60 - 45 = 15 2 ) b is one fifth of the sum of a and c or b = 1 / 5 ( a + c ) . . . but a + b + c = 60 so 1 / 5 ( a + c ) + a + c = 60 a + c = 60 * 5 / 6 = 50 but a = 15 , so c = 50 - 15 = 35 answer : c
Answer: c
Problem: the instructions state that cheryl needs 5 / 11 square yards of one type of material and 2 / 3 square yards of another type of material for a project . she buys exactly that amount . after finishing the project , however , she has 25 / 55 square yards left that she did not use . what is the total amount of square yards of material cheryl used ?
Options: a ) 1 / 12 , b ) 1 / 9 , c ) 1 2 / 3 , d ) 2 / 3 , e ) 2 1 / 9
Rationale: "total bought = 5 / 11 + 2 / 3 left part 25 / 55 - - - > 5 / 11 so used part 5 / 11 + 2 / 3 - 5 / 11 = 2 / 3 ans d"
Answer: d
Problem: in a class of 87 students 41 are taking french , 22 are taking german . of the students taking french or german , 9 are taking both courses . how many students are not enrolled in either course ?
Options: a ) 6 , b ) 15 , c ) 24 , d ) 33 , e ) 54
Rationale: formula for calculating two overlapping sets : a + b - both + not ( a or b ) = total so in our task we have equation : 41 ( french ) + 22 ( german ) - 9 ( both ) + not = 87 54 + not = 87 not = 87 - 54 = 33 so answer is d
Answer: d
Problem: what is the least positive integer x for which ( x ^ 2 ) + 164 is divisible by 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , and 11 ?
Options: a ) 27556 , b ) 800 , c ) 796 , d ) 87 , e ) 166
Rationale: 2 = 2 3 = 3 4 = 2 * 2 5 = 5 6 = 2 * 3 7 = 7 8 = 2 * 2 * 2 9 = 3 * 3 10 = 2 * 5 11 = 11 so ( x ^ 2 ) + 164 = ( 2 * 2 * 2 * 3 * 3 * 5 * 7 * 11 ) = 27720 . then ( x ^ 2 ) = 27556 . then x = 166 , which is option e .
Answer: e
Problem: the cross - section of a cannel is a trapezium in shape . if the cannel is 16 m wide at the top and 4 m wide at the bottom and the area of cross - section is 700 sq m , the depth of cannel is ?
Options: a ) 76 , b ) 28 , c ) 27 , d ) 80 , e ) 70
Rationale: "1 / 2 * d ( 16 + 4 ) = 700 d = 70 answer : e"
Answer: e
Problem: two equal triangles constitute rhombus a . if the diagonals of rhombus are 15 cm and 20 cm long . what is the area of each triangle ?
Options: a ) 300
b ) 150
c ) 75
d ) 40
e ) 20
Rationale: area of rhombus = 0.5 x 15 x 20 = 150 cm â ² area of triangle = 150 / 2 = 75 cm â ² answer : c
Answer: c
Problem: there is a total of 84 marbles in a box , each of which is red , green , blue , or white . if one marble is drawn from the box at random , the probability that it will be white is 1 / 4 and the probability that it will be green is 1 / 7 . what is the probability that the marble will be either red or blue ?
Options: a ) 4 / 7 , b ) 9 / 14 , c ) 11 / 14 , d ) 17 / 28 , e ) 23 / 28
Rationale: "p ( red or blue ) = 1 - p ( white ) - p ( green ) = 28 / 28 - 7 / 28 - 4 / 28 = 17 / 28 the answer is d ."
Answer: d
Problem: if the side length of square b is four times that of square a , the area of square b is how many times the area of square a ?
Options: a ) 20
b ) 16
c ) 12
d ) 8
e ) 4
Rationale: let x be the side length of square a . then the area of square a is x ^ 2 . the area of square b is ( 4 x ) ^ 2 = 16 x ^ 2 . the answer is b .
Answer: b
Problem: two trains of equal are running on parallel lines in the same direction at 50 km / hr and 36 km / hr . the faster train passes the slower train in 36 sec . the length of each train is ?
Options: a ) 70 m , b ) 72 m , c ) 80 m , d ) 82 m , e ) 84 m
Rationale: "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 50 - 36 = 14 km / hr . = 14 * 5 / 18 = 35 / 9 m / sec . 2 x / 36 = 35 / 9 = > x = 70 . answer : a"
Answer: a
Problem: a dog takes 3 leaps for every 7 leaps of a hare . if one leap of the dog is equal to 3 leaps of the hare , the ratio of the speed of the dog to that of the hare is :
Options: a ) 9 : 7 , b ) 9 : 9 , c ) 9 : 6 , d ) 9 : 1 , e ) 9 : 2
Rationale: "explanation : dog : hare = ( 3 * 3 ) leaps of hare : 7 leaps of hare = 9 : 7 . answer : a ) 9 : 7"
Answer: a
Problem: what is the dividend . divisor 17 , the quotient is 9 and the remainder is 5 ?
Options: a ) 150 , b ) 152 , c ) 154 , d ) 158 , e ) 144
Rationale: "d = d * q + r d = 17 * 9 + 5 d = 153 + 5 d = 158"
Answer: d
Problem: circular gears p and q start rotating at the same time at constant speeds . gear p makes 10 revolutions per minute and gear q makes 40 revolutions per minute . how many seconds after the gears start rotating will gear q have made exactly 2 more revolutions than gear p ?
Options: a ) 4 , b ) 8 , c ) 10 , d ) 12 , e ) 15
Rationale: "gear q makes 30 more revolutions each 60 seconds . the time to make 2 = 30 / 15 more revolutions is 60 / 15 = 4 seconds . the answer is a ."
Answer: a
Problem: a and b together can do a piece of work in 3 days . if a alone can do the same work in 20 days , then b alone can do the same work in ?
Options: a ) 0.28 days , b ) 0.45 days , c ) 0.55 days , d ) 0.25 days , e ) 0.15 days
Rationale: "b = 1 / 2 – 1 / 20 = 0.28 days answer : a"
Answer: a
Problem: the hcf of two numbers is 62 and the other two factors of their lcm are 11 and 12 . what is the largest number .
Options: a ) 462 , b ) 744 , c ) 480 , d ) 504 , e ) 555
Rationale: "explanation : hcf of the two numbers = 62 hcf will be always a factor of lcm 62 is factor of lcm other two factors are 11 & 12 then the numbers are ( 62 * 11 ) and ( 62 x 12 ) = 682 and 744 answer : option b"
Answer: b
Problem: a and b can do a piece of work in 45 days and 40 days respectively . they began to do the work together but a leaves after some days and then b completed the remaining work in 23 days . the number of days after which a left the work was ?
Options: a ) 12 , b ) 11 , c ) 10 , d ) 9 , e ) 8
Rationale: let the total units of work to be done be 360 . the units of work done by a in a single day = 8 similarly , the units of work done by b in a single day = 9 . a and b ’ s one day work = 17 units a and b worked together for some days = 17 x ( assume ) b ’ s work alone for 23 days = 23 x 9 = 207 so , the work done by a and b together = ( 360 - 207 ) = 153 units therefore , 17 x = 153 = > 9 units therefore , the number of days after which a left the work was 9 days . answer : d
Answer: d