RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: A can do a piece of work in 10 days and B alone can do it in 20 days. How much time will both take to finish the work ?
A.A)5.333
B.B)6
C.C)6.666
D.D)8.333
Answer: C.C)6.666
Explanation: This question can be solved by different methods. We need to conserve time in exams so solving this problem using equations is the good idea.
Time taken to finish the job = XY / (X + Y)
= 10 x 20 / (10 + 20)
= 200 / 30
= 6.666 days
Answer: C
Question: ----------------YES---------NO----UNSURE
Subject M----500--------200-----100
Subject R----400--------100-----300
A total of 800 students were asked whether they found two subjects, M and R, interesting. Each answer was either yes or no or unsure, and the numbers of students who gave these answers are listed in the table above. If 180 students answered yes only for subject M, how many of the students did not answer yes for either subject?
A.100
B.220
C.300
D.400
Answer: B.220
Explanation: Since 180 students answered yes only for subject M, then the remaining 320 students who answered yes for subject M, also answered yes for subject R. So, 320 students answered yes for both subjects.
If 320 students answered yes for both subjects, then 400-320=80 students answered yes only for subject R.
So, we have that:
200 students answered yes only for subject M;
80 students answered yes only for subject R;
300 students answered yes for both subjects;
Therefore 800-(200+80+300)=220 students did not answer yes for either subject.
Answer: B.
Question: when i was married 10 years ago,my wife was the 6th member in the family.today,my father died,and a baby was born to me.the average age in my family today is what it was when i was married.how old was my father when he died?
A.50yrs
B.60yrs
C.70yrs
D.80yrs
Answer: B.60yrs
Explanation: 10 years ago the family members there were 6 members with age= I+Father+Wife+3 others
Now 6 members with age= (I+10)+(Wife +10)+ (3 others+3*10) + (baby=0).
I+Father+Wife+3 others=(I+10)+(Wife +10)+ (3 others+3*10) + (baby=0)
Father =50+0 = 50 = Father's age 10 year ago
So now father's age=50 +10=60.
ANSWER:B
Question: A can do a piece of work in 7 days and B can do it in 6 days how long will they both work together to complete the work?
A.6/11
B.8/13
C.3/13
D.9/11
Answer: C.3/13
Explanation: Explanation:
A’s one day work = 1/5
B’s one day work = 1/6
(A+B)’s one day work = 1/7 + 1/6 = 13/42 => time = 42/13 = 3 3/13 days
Answer: Option C
Question: At a certain university, 65% of the professors are women, and 70% of the professors are tenured. If 90% of the professors are women, tenured, or both, then what percent of the men are tenured?
A.25
B.37.5
C.50
D.62.5
Answer: D.62.5
Explanation: total women = 65%
total men = 40%
total tenured = 70% (both men and women)
therefore, women tenured + women professors + men tenured = 90%
men tenured = 25%
but question wants to know the percent of men that are tenured
25%/40% = 62.5%
D
Question: The marks obtained by Vijay and Amith are in the ratio 4:5 and those obtained by Amith and Abhishek in the ratio of 3:2. The marks obtained by Vijay and Abhishek are in the ratio of?
A.6:8
B.6:2
C.6:5
D.6:4
Answer: C.6:5
Explanation: 4:5
3:2
-------
12:15:10
12:10
6:5
Answer: C
Question: Rs.2500 is divided into two parts such that if one part be put out at 5% simple interest and the other at 6%, the yearly annual income may be Rs.145. How much was lent at 5%?
A.2333
B.2777
C.2688
D.500
Answer: D.500
Explanation: (x*5*1)/100 + [(2500 - x)*6*1]/100 = 145
X = 500
Answer: D
Question: Calculate the number of bricks, each measuring 25 cm x 15 cm x 8 cm required to construct a wall of dimensions 10 m x 4 m x 5 m when 10% of its volume is occupied by concrete ?
A.6000
B.5400
C.3800
D.4700
Answer: A.6000
Explanation: Explanation:
Let 'B' be the nuber of bricks.
=> 10 x 4/100 x 5 x 90/100 = 25/100 x 15/100 x 8/100 x B
=> 10 x 20 x 90 = 15 x 2 x B
=> B = 6000
ANSWER IS A
Question: A number is said to be prime saturated if the product of all the different positive prime factors of g is less than the square root of g. What is the greatest two digit prime saturated integer ?
A.99
B.98
C.97
D.96
Answer: D.96
Explanation: g=96 = 3 * 32 = 3 * 2^5
Answer is D.
Question: David and Rachel are getting married. The extended family wants to have its picture taken, but David's father is unwilling to stand next to Rachel's mother. How many options does the photographer have to arrange the 13 family members in a row for the picture?
A.14!
B.12×11!
C.11×12!
D.13!/2!
Answer: C.11×12!
Explanation: The number of ways to arrange 13 people is 13!
We need to subtract the arrangement when these two people are together.
Let's think of these two people as a single unit so there are 12 units.
The number of ways to arrange 12 units is 12!
We need to multiply this by 2 since these two people could switch places.
The total number of valid arrangements is 13! - 2*12! = 12!*(13-2) = 11x12!
The answer is C.
Question: At a certain stage of a soccer tournament, the score ratio of teams A, B and C was 3:4:5. Eventually, the score ratio of A to B has doubled while the score ratio of A to C has halved. If the final score of team C was 40, what was the final score A of team B?
A.1)8
B.2)10
C.3)20
D.4)40
Answer: A.1)8
Explanation: A to B = 3 : 4
So, on doubling we get 6 : 4
A to C = 3 : 5
So, on halving we get 1.5 : 5 or 3 : 10 or 6 : 20
So final ratio = 6 : 4 : 20.
If 20x = 40
4x = 8
Hence, answer is A
Question: a is the hundreds digit of the three digit integer x, b is the tens digit of x, and c is the units digit of x. a = b = 3c, and a > 0.
What is the difference between the two greatest possible values of x?
Tip: Dont stop till you have exhausted all answer choices to arrive at the correct one.
A.112
B.213
C.331
D.511
Answer: C.331
Explanation: Ratio of a:b:c= 3:3:1
Two possible greatest single digit values for a are 9 and 6
if a is 9, then x= 993
if a is 6, then x= 662
difference= 993-662= 331
C is the answer
Question: . A car covers a distance of 810 km in 5 hours. Find its speed?
A.104
B.55
C.162
D.174
Answer: C.162
Explanation: 810/5 = 162 kmph
Answer: C
Question: Set E consists of all the prime numbers between 10 and 25. What is the range of set E?
A.12
B.13
C.16.6
D.17
Answer: A.12
Explanation: Answer of set E is 12=A
Question: If 7<x<9<y<15, then what is the greatest possible positive integer difference of x and y?
A.7
B.4
C.5
D.6
Answer: A.7
Explanation: Let x=7.1 and y=14.1
Greatest possible difference=14.1-7.1=7
Answer A
Question: The average age of 32 students in a group is 10 years. when teacher's age is included to it, the average increases by one.What is the teacher's age in years?
A.45
B.43
C.51
D.53
Answer: B.43
Explanation: age of the teacher=(33*11-32*10)
=43 years.
ANSWER:B
Question: Three candidates contested in an election and received 2136, 7636, 11628 votes respectively. What is the percentage of the total votes did the winning candidate get?
A.45%
B.49%
C.57%
D.54%
Answer: D.54%
Explanation: Total number of votes polled = 2136+7636+11628 = 21400
Required percentage = 11628*100/21400 = 54%
Answer is D
Question: If 5x = 6y and xy ≠ 0, what is the ratio of 1/3*x to 1/5*y ?
A. 25/6
B. 36/25
C. 2
D. 5/6
Answer: C. 2
Explanation: 5x = 6y => x/y = 6/5
1/3*x to 1/5*y = x/y * 5/3 = (6/5)*(5/3)=2
Ans: C
Question: Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
A.7 min
B.8 min
C.9 min
D.10 min
Answer: D.10 min
Explanation: Due to stoppages, it covers 9 km less.
Time taken to cover 9 km =(9/54)x60 = 10 min
answer : D
Question: In an examination the number of successful students was 4 times the numbr of unsuccessful students If there were 35 less students nd the numbr of unsuccessful students were 9 more,then the ratio of their numbr would have been 2:1 . Find the total numbr of students
A.155
B.145
C.135
D.125
Answer: A.155
Explanation: successful students 4x,unsuccessful student x.total 5x
now,total students become 5x-35 & unsuccessful students x+9 then successful students 4x-44
ratio 4x-44 / x+9=2 / 1
x=31 then total student 5x=155
ANSWER:A
Question: 5 OUT OF 1000 PARTS OF EARTH IS SULPHUR. WHAT IS THE PERCENTAGE OF SULPHUR IN EARTH
A.11/50
B.2/9
C.1/2
D.2/45
Answer: C.1/2
Explanation: REQUIRED PERCENTAGE=(5/1000*100)%=1/2%
CORRECT OPTIONS: C
Question: For any number d, d* is defined as the greatest positive even integer less than or equal to d. What is the value of 5.2 – 5.2*?
A.0.2
B.1.2
C.1.8
D.2.2
Answer: B.1.2
Explanation: Since d* is defined as the greatest positive even integer less than or equal to d, then 5.2* = 4 (the greatest positive even integer less than or equal to 5.2 is 4).
Hence, 5.2 – 5.2* = 5.2 - 4 = 1.2
Answer: B.
Question: On a clock, how many times a day do the minute and hour hand overlap?
A.10
B.11
C.20
D.22
Answer: D.22
Explanation: 22 times: 12:00:00, 1:05:27, 2:10:55, 3:16:22, 4:21:49, 5:27:16, 6:32:44, 7:38:11, 8:43:38, 9:49:05, 10:54:33. Each twice a day.
Question: The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is
A.22
B.40
C.38
D.21
Answer: B.40
Explanation: Explanation:
Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years.
Sum of the present ages of wife and child (20 x 2 + 5 x 2) years = 50 years.
Husband's present age = (90 - 50) years = 40 years.
Answer: B
Question: If the cost of A metres of wire is C rupees, then what is the cost of E metres of wire at the same rate?
A.Rs. (EC/A)
B.Rs. (EC)
C.Rs. (C/A)
D.Rs. (E/A)
Answer: A.Rs. (EC/A)
Explanation: cost of A metres of wire = Rs.C
cost of 1 metre of wire = Rs.(C/A)
cost of E metre of wire = Rs.(E×C/A)=Rs. (EC/A)
ANSWER:A
Question: Which one of the following is not a prime number?
A.31
B.61
C.71
D.91
Answer: D.91
Explanation: Explanation:
91 is divisible by 7. So, it is not a prime number.
ANSWER IS D
Question: Let y = 2g + x^2 and g = 5x + 1.
If g^2 = 121, then which of the following can be a value of 2y + 3g?
A.81
B.130
C.141
D.142
Answer: A.81
Explanation: 800score Official Solution:
First note that g can either be 11 or -11. We are going to have to use both to find all solutions before we can choose a final answer. We begin by substituting g = 11 and then g = -11. If g = 11, then x = 2. If g = -11, then x = -12/5. Since all of our solutions are whole numbers, we can dismiss -12/5. We use g = 11 and x = 2 to determine the value of y: y = 2(11) + 4 = 24. Finally, we substitute the values for y and g into the last expression to determine its value : 2(24) + 3(11) = 48 + 33 = 81;the correct answer is (A).
Question: If Jan 1, 2006 was a Monday, What was the day of the week Jan 1, 2010?
A.Friday
B.Thursday
C.Tuesday
D.Saturday
Answer: D.Saturday
Explanation: Explanation :
Given that Jan 1 2006 was a Monday
Number of odd days in the period 2006-2009
= 3 normal years + 1 leap year
= 3 x 1 + 1 x 2 = 5 (note that we have taken the complete year 2006 because the period in 2006 is from 01-Jan-2006 to 31-Dec-2006, which is the whole year 2006. Then the complete years 2007, 2008 and 2009 are also involved)
Hence Jan 1 2010 = (Monday + 5 odd days) = Saturday.. Answer : Option D
Question: The present ratio of students to teachers at a certain school is 50 to 1. If the student enrollment were to increase by 50 students and the number of teachers were to increase by 5, the ratio of students to teachers would then be 25 to 1. What is the present number of teachers?
A.3
B.8
C.10
D.12
Answer: A.3
Explanation: We are given that the ratio of students to teacher is 50 to 1. We can rewrite this using variable multipliers.
students : teachers = 50x : x
We are next given that student enrollment increases by 50 and the number of teachers increases by 5. With this change the new ratio becomes 25 to 1. We can put all this into an equation:
Students/Teachers  25/1 = (30x + 50)/(x + 5)
If we cross multiply we have:
25(x + 5) = 50x + 50
25x + 125 = 50x + 50
3 = x
Since x is the present number of teachers, currently there are 3 teachers.
Answer A.
Question: A man can hit the target once in two. If he fires four shots in succession, what is the probability that he will hit the target?
A.1
B.1/16
C.81/16
D.15/16
Answer: D.15/16
Explanation: In four shots he can hit once,twice,thrice,all hit
the probability of hitting the target is p(1hit out of 4)+P(2hit out of 4)+p(3hit out of 4)+p(All hit)
it is total probability-probability of not hitting the target
=>1-(1/2*1/2*1/2*1/2)
=>15/16
ANSWER:D
Question: A certain ball team has an equal number of right- and left-handed players. On a certain day, two-thirds of the players were absent from practice. Of the players at practice that day, two-third were left handed. What is the ratio of the number of right-handed players who were not at practice that day to the number of lefthanded players who were not at practice?
A.1/3
B.2/3
C.5/7
D.7/5
Answer: D.7/5
Explanation: Say the total number of players is 18, 9 right-handed and 9 left-handed.
On a certain day, two-thirds of the players were absent from practice --> 12 absent and 6 present.
Of the players at practice that day, one-third were left-handed --> 6*2/3=4 were left-handed and 2 right-handed.
The number of right-handed players who were not at practice that day is 9-2=7.
The number of left-handed players who were not at practice that days is 9-4=5.
The ratio = 7/5.
Answer: D
Question: If x is chosen at random from the set {1, 2, 3, 4} and y is to be chosen at random from the set {5, 6, 7}, then what is the probability that xy will be even?
A.5/6
B.2/6
C.2/3
D.1/2
Answer: C.2/3
Explanation: S = {(1,5), (1,6), (1,7), (2,5), (2,6), (2,7), (3,5), (3,6), (3,7), (4,5), (4,6), (4,7)}
Total element n(S) = 12
xy will be even when even x or y or both will be even.
Events of xy being even is E.
E = {(1,6), (2,5), (2,6), (2,7), (3,6), (4,5), (4,6), (4,7)}
n(E) = 8
P(E) = n(E)/n(S)
= 8/12
= 2/3
ANSWER:C
Question: If 5% more is gained by selling an book for $ 350 than by selling it for $ 340, the cost of the book is:
A.500
B.400
C.200
D.300
Answer: C.200
Explanation: C
200
Let C.P. be $ x.
Then, 5% of x = 350 - 340 = 10
x/20 = 10 => x = 200
Question: If 0 < x ≤ 4, then which one of the following is the maximum value of (x – 1)^2 + x ?
A.13
B.–1
C.0
D.1
Answer: A.13
Explanation: the maximum value of (x – 1)^2 + x for 0 < x ≤ 4 is when x=4 --> (x – 1)^2 + x = 13
Answer A
Question: Three persons invested Rs.9000 in a joint business. The second person invested Rs.1000 more than the first and the third Rs.1000 more than second. After two years, they gained Rs.5400. How much third person will get?
A.2400
B.2429
C.2428
D.2421
Answer: A.2400
Explanation: First persons investment = x
Second persons investment = x + 1000
Third persons investments = x + 2000
x + x + 1000 + x + 2000 = 9000
3x = 6000
x = 2000
Ratio = 2000 : 3000 : 4000
2:3:4
4/9 * 54000 = 2400
Answer:A
Question: A rectangular lawn of dimensions 80 m * 60 m has two roads each 10 m wide running in the middle of the lawn, one parallel to the length and the other parallel to the breadth. What is the cost of traveling the two roads at Rs.3 per sq m?
A.3929
B.3998
C.3128
D.3900
Answer: D.3900
Explanation: Area = (l + b – d) d
(80 + 60 – 10)10 => 1300 m2
1300 * 3 = Rs.3900
Answer: D
Question: If 1/a^3 + a^3 represents the diameter of circle O and 1/a + a = 3, which of the following best approximates the circumference of circle O?
A.28
B.22
C.20
D.16
Answer: C.20
Explanation: Given that (1/a)+a =3
Square both sides of the equation:
We get,
[(1/a)^2 + a^2 + 2*(1/a)*a] = 9
=> (1/a)^2 + a^2 + 2 = 9
=> (1/a)^2 + a^2 = 7 ----------------- (1)
Diameter D = (1/a)^2 + a^2
= 7 (From (1))
So Radius = D/2 = 7/2
Circumference = 2*Pi*r
= 2*(22/7)*(7/2)
=20
So the answer should be C
Question: The curved surface area of a cylindrical pillar is 264 sq.m and its volume is 924 cub.m. Find the ratio of its diameter to its height
A.7:9
B.7:3
C.7:7
D.7:6
Answer: B.7:3
Explanation: Explanation:
{\color{Black} \frac{\prod r^{2}h}{2\prod rh}=\frac{924}{264}}
{\color{Black} \Rightarrow r=\frac{924}{264}\times 2=7m}
{\color{Black}And\; \; 2\prod rh =264 \Rightarrow h=(264\times \frac{7}{22}\times \frac{1}{2}\times \frac{1}{7})=6m}
{\color{Black}\therefore } Required Rato ={\color{Black}\frac{2r}{h} } ={\color{Black}\frac{14}{6}=7:3 }
Answer: B) 7 : 3
Question: In how many ways can 5 GIRLS and 3 BOYS be seated on 8 chairs so that no two BOYS are together?
A.A)5760
B.B)14400
C.C)480
D.D)56
Answer: B.B)14400
Explanation: Consider the following arrangement:
*G*G*G*G*G
Now, if BOYS occupy the places of 6 stars no BOYS will be together.
# of ways 3 BOYS can occupy the places of these 6 stars is C36=20;
# of ways 3 BOYS can be arranged on these places is 3!=6
# of ways 5 GIRLS can be arranged is 5!=120
So total # of ways to arrange 3 BOYS and 5 GIRLS so that no BOY are together is 20∗6∗120=14,400
ANS:B
Question: S is a set of 45 consecutive multiples of 5. If the smallest number in S is 55, then the greatest number in S is
A.198
B.275
C.299
D.402
Answer: B.275
Explanation: last term =first term +(total no. of terms-1)consecutive difference
S is a set of 45 consecutive multiples of 5. If the smallest number in S is 55, then the greatest number in S is
first term = 55 ; total terms =45 ; difference = 5
55+(44)5 = 275
Ans B
Question: A salesman commission is 10% on all sales upto $5000 and 5% on all sales exceeding this. He remits $15000 to his parent company after deducting his commission. Find the total sales?
A.15456
B.14758
C.15789
D.13250
Answer: C.15789
Explanation: Let his total sales be x
Total sales - commission = $15000
x - [(10% of 5000)+5% of (x-5000)] = 15000
95x/100 = 15000
x = 15789 approximately
Answer is C
Question: The product of all the prime numbers less than 24 is closest to which of the following powers of 10?
A.10^9
B.10^8
C.10^7
D.10^6
Answer: B.10^8
Explanation: OA: B
Question: If the cost price of 12 pens is equal to the selling price of 11 pens, the gain percent is:
A.80%
B.90%
C.9%
D.40%
Answer: C.9%
Explanation: Explanation:
Let C.P. of each pen be Re. 1.
Then, C.P. of 11 pens = Rs. 11; S.P. of 11 pens = Rs. 12.
Gain % = 1/11 * 100 = 9%
Answer:C
Question: One boy can eat 100 chocolates in half a minute, and another can eat half as many in twice the length of time. How many chocolates can both boys eat in 15 seconds?
A.61.5 chocolates
B.63.5 chocolates
C.62.5 chocolates
D.64.5 chocolates
Answer: C.62.5 chocolates
Explanation: 1st boy eats 100 chocos / 30 sec
so he can eat 50 in 15 sec
2nd boy can eat one-half of 100 in twice of 30 sec
so he can eat 50 in 60 sec
or that is 12.5 in 15 sec
so, together they can eat
50 + 12.5 or
62.5 chocolates in 15 seconds.
ANSWER:C
Question: Robert is traveling on his cycle and has calculated to reach point A at 2 p.m. if he travels at 10 km/hr; he will reach there at 12 noon if he travels at 15 km/hr. At what speed must he travel to reach A at 1 p.m.?
A.12 kmph
B.18 kmph
C.19 kmph
D.15 kmph
Answer: A.12 kmph
Explanation: Let the distance traveled be x km.
Then, x/10 - x/15 = 2
3x - 2x = 60 => x = 60 km.
Time taken to travel 60 km at 10 km/hr = 60/10 = 6 hrs.
So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m.
Required speed = 60/5
=12 kmph.
Answer:A
Question: If 28^2*17^3*12^2 = x, what is the units digit of x ?
A.2
B.8
C.6
D.7
Answer: B.8
Explanation: 28^2*17^3*12^2 = x
Units digit of x will be
8^2 * 7^3 * 2^2
=2* 3 * 4
=8
Therefore , units digit of x = 8
Answer B
Question: Given three boxes I, II and III, each containing two coins. In box I, both coins are 1 rupee
coins, in box II, both are 2 rupee coins and in the box III, there is one 1 rupee and one 2 rupee coin.
A person chooses a box at random and takes out a coin. If the coin is of 1 rupee, what is the
probability that the other coin in the box is also of 1 rupee?
A.1/5
B.2/3
C.2/7
D.3/7
Answer: B.2/3
Explanation: Let E1, E2 and E3 be the events that boxes I, II and III are chosen, respectively.
Then P(E1) = P(E2) = P(E3) = 1/3
Also, let A be the event that ‘the coin drawn is of 1 rupee'
Then P(A|E1) = P(a 1 rupee coin from bag I) = 2/2 = 1
P(A|E2) = P(a 1 rupee coin from bag II) = 0
P(A|E3) = P(a 1 rupee coin from bag III) = 1/2
the probability that the other coin in the box is of 1 rupee
= the probability that 1 rupee coin is drawn from the box I.
= P(E1|A)
from Bayes' theorem, we know that
P(E1|A) = [P(E1 )P(A|E1 )]/ [P(E1 )P(A|E1 )+P(E2 )P(A|E2 )+P(E3 )P(A|E3 )]
= (1/3 × 1)/(1/3 × 1 + 1/3 × 0 + 1/3 × 1/2)
= (1/3)/(1/3 × 1 + 1/3 × 0 + 1/3 × 1/2)
= (1/3)/(1/3 × 1/6) = (1/3)/(3/6) = (1/3)/2 = 2/3
B
Question: The difference between a number and its two-fifth is 510. What is 20% of that number?
A.58
B.170
C.95
D.100
Answer: B.170
Explanation: Let the number be x. Then,
x - 2/5 x = 510
x = (510 * 5)/3 = 850
20% of 850 = 170.
Answer:B
Question: Seats for Mathematics, Physics and Biology in a school are in the ratio 5:7:8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?
A.2:3:4
B.2:3:8
C.2:3:0
D.2:3:1
Answer: A.2:3:4
Explanation: Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively. Number of increased sears are (140% of 5x), (150% of 7x) and (175% of 8x)
i.e., (140/100 * 5x), (150/100 * 7x) and (175/100 * 8x)
i.e., 7x, 21x/2 and 14x
Required ratio = 7x:21x/2:14x
= 14x : 21x : 28x = 2:3:4
Answer: A
Question: 36 men can complete a piece of work in 18 days. in how many days will 27 men complete the same work ?
A.12
B.18
C.22
D.24
Answer: D.24
Explanation: Solution
Let the required number of days be x.
Then, Less men, More days
∴ 27 : 36 :: 18 : x ⇔ 27 × x = 36 × 18 ⇔ x =36x18/27x = 24. Answer D
Question: A standard Veggiematik machine can chop 36 carrots in 4 minutes. How many carrots can 6 standard Veggiematik machines chop in 5 minutes?
A.36
B.54
C.108
D.270
Answer: D.270
Explanation: ONE Veggiematik machine can chop 36 carrots in 4 minutes
So, ONE Veggiematik machine can chop (1.5)(36) carrots in (1.5)(4) minutes
In other words, ONE Veggiematik machine can chop 54 carrots in5minutes
So,SIXVeggiematik machines can chop (5)(54) carrots in5minutes
In other words, SIX Veggiematik machines can chop 270 carrots in5minutes
Answer D
Question: Two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 16 kmph and 19 kmph respectively. When they meet, it is found that one train has traveled 60 km more than the other one. The distance between the two stations is?
A.565
B.444
C.676
D.420
Answer: D.420
Explanation: 1h ----- 3
? ------ 60
12 h
RS = 16 + 19 = 35
T = 12
D = 35 * 12 =420
Answer: D
Question: A cubical tank is filled with water to a level of 2 feet. If the water in the tank occupies 32 cubic feet, to what fraction of its capacity is the tank filled with water?
A.1/2
B.1/3
C.2/3
D.1/4
Answer: A.1/2
Explanation: The volume of water in the tank is H*L*B = 32 cubic feet.
Since H = 2, then L*B = 16 and L = B = 4.
Since the tank is cubical, the capacity of the tank is 4*4*4 = 64.
The ratio of the water in the tank to the capacity is 32/64 = 1/2
The answer is A.
Question: A rower can row 7 km/h in still water. When the river is running at 2 km/h, it takes the rower 1 hour to row to Big Rock and back. How many kilometers is it to Big Rock?
A.2.92
B.3.04
C.3.12
D.3.21
Answer: D.3.21
Explanation: Let x be the distance to Big Rock.
time = x/5 + x/9 = 1
x = 45/14 = 3.21 km
The answer is D.
Question: A can do a piece of work in 4 hours . A and C together can do it in just 2 hours, while B and C together need 3 hours to finish the same work. B alone can complete the work in --- hours.
A.65
B.85
C.48
D.12
Answer: D.12
Explanation: Explanation :
Work done by A in 1 hour = 1/4
Work done by B and C in 1 hour = 1/3
Work done by A and C in 1 hour = 1/2
Work done by A,B and C in 1 hour = 1/4+1/3 = 7/12
Work done by B in 1 hour = 7/12 – 1/2 = 1/12 => B alone can complete the work in 12 hours
ANSWER:D
Question: The salary of Mr. X is 30% more than that of Mr. Y . Find what percent of Mr. Y’s salary is less than Mr. X’s?
A.30%
B.25 1/13 %
C.23 1/13 %
D.22 1/13 %
Answer: C.23 1/13 %
Explanation: y=100	x=130
130---------30
100---------? => 23 1/3%
ANSWER:C