mathewhe/aqua_rat huggingface dataset:
Question: If the LCM of two integers c, d (where d> c and c>1) is c*d, then which of the following can not be true?
(A) Both c and d can be consecutive integers.
(B) Both c and d can be Prime Numbers.
(C) All Prime factors of c must be Prime factors of d.
(D) c and d do not have a common factor.
(E) c and d do not share any Prime factors.
Rationale: If LCM of two numbers C and D is the product CD then both numbers are Co- prime.
(Co-prime numbers are any two numbers which have an HCF of 1 i.e. when two numbers have no common prime factor apart from the number 1).
Numbers that can be Co-Prime are :
1) Two consecutive natural numbers.
2) Two consecutive odd numbers.
3) Two prime numbers.
4) One prime number and the other a composite number such that the composite number is not a multiple of the prime number.
With above four statements it can be concluded that Options A, B, D and E are all true.
Only Option C cannot be true.
Hence answer is C
Answer: (C) All Prime factors of c must be Prime factors of d.
Question: Vishal invested 10% more than Trishul. Trishul invested 10% less than Raghu. If the total sum of their investments is Rs. 6647, how much amount did Raghu invest ?
(A) 2300
(B) 2887
(C) 2000
(D) 1129
(E) 1192
Rationale: Let money invested by Raghu = Rs. x
Money invested by Trishul = 9/10 x = 0.9x
Money invested by Vishal = 9/10x * 110/100 = 0.99x
Also, x+0.9x+0.99x = 6647
= x= 6647/2.89 = 2300
Therefore, amount invested by Raghu is Rs. 2300.
Answer: A
Answer: (A) 2300
Question: At what time between 9 and 10 o’clock will the hands of a watch be together?
(A) 49 1/11 min. past 9
(B) 40 min. past 9
(C) 45 min. past 9
(D) 50 min. past 9
(E) None
Rationale: Sol.
To be together between 9 and 10 o’clock, the minute hand has to gain 45 min. spaces. 55 min. spaces gined in 60 min.
45 min. spaces are gained in [60 / 55 * 45] min. or 49 1/11 min.
∴ The hands are together at 49 1/11 min. past 9.
Answer A
Answer: (A) 49 1/11 min. past 9
Question: A sum of Rs. 427 is to be divided among A, B and C such that 3 times A’s share, 4 tunes B’s share and 7 times C’s share are all equal. The share of C is :
(A) 84
(B) 140
(C) 140
(D) 240
(E) None of these
Rationale: Explanation:
3A = 4B = 7C = k,Then A = k/3, B = k/4 and C= k/7.
A : B : C = k/3 : k/4 : k/7 = 28:21 :12.
Cs share = Rs. [427 x (12/61)] = Rs. 84
Answer: A
Answer: (A) 84
Question: The weights of one liter vegetable ghee packet of two brands ‘A’ and ‘B’ are 900 gm and 850 gm respectively. If they are mixed in the ratio of 3 : 2 by volumes to form a mixture of 4 liters, what is the weight (in kg) of the mixture?
(A) 3.84
(B) 1.75
(C) 3.52
(D) 2.72
(E) None of these
Rationale: 1L A = 900 gm
1L B = 850 gm
We are mixing five parts (3 parts A plus 2 parts B, 5 parts total) to get 4L, so 5x = 4 ---> x = 4/5. Each "part" is 4/5 of a liter.
So if we have 3 parts A, we have 900*3*(4/5) = 2160
If we have 2 parts B, we have 850*2*(4/5) = 1360
2160 + 1360 = 3520
Solving for units gives us 3.52
So the answer is C
Answer: (C) 3.52
Question: What profit percent is made by selling an article at a certain price, if by selling at 2/3rd of that price, there would be a loss of 10%?
(A) 20%
(B) 35%
(C) 13 1/30%
(D) 12%
(E) 13%
Rationale: SP2 = 2/3 SP1
CP = 100
SP2 = 90
2/3 SP1 = 90
SP1 = 135
100 --- 35 => 35%
ANSWER:B
Answer: (B) 35%
Question: While planning their outing, Abhishek understood that their boat could travel with a speed of 12kmph in still water.The group decided to travel 24 minutes downstream. Aarti had prior experience in that route and mentioned that the speed of stream was 4kmph.How many kms will the group travel downstream?
(A) 6.4
(B) 3.2
(C) 4
(D) 4.8
(E) 4.6
Rationale: downstream speed = speed of the boat + speed of the river
12+4 = 16
distance travelled = 16x24/60 = 6.4
ANSWER:A
Answer: (A) 6.4
Question: What will come in place of the x in the following Number series? 46080, 3840, 384, ?, 8, 2, 1
(A) 1
(B) 3
(C) 5
(D) 48
(E) 9
Rationale: 46080 /12 = 3840
3840 /10 = 384
384 /8 = 48
48 /6 = 8
8/4 = 2
2 /2 = 1
D
Answer: (D) 48
Question: What is the solution of the equations x - y = 0.08 and 1(x + y)-1 = 2 ?
(A) x = 3.2, y = 2.3
(B) x = 1, y = 0.1
(C) x = 2, y = 1.1
(D) x = 0.065, y = 0.735
(E) None
Rationale: Answer
x - y = 0.8 ...(i)
and 1(x + y)-1=2
⇒ 1/ (x + y) = 2
⇒ 2(x + y) =1
⇒ x + y = 1/2 ...(ii)
On solving Eqs.(i) and (ii),we get
x = 0.065
and y = 0.735
Correct Option: D
Answer: (D) x = 0.065, y = 0.735
Question: John purchased a grinder and a mobile for Rs. 15000 &Rs. 10000 respectively. He sold the grinder at a loss of 4% and the mobile phone at a profit of 10%. Overall how much he make a profit.
(A) s. 400
(B) s. 120
(C) s. 420
(D) s. 450
(E) s. 290
Rationale: Let the SP of the refrigerator and the mobile phone be Rs. r and Rs. m respectively.
r = 15000(1 - 4/100) = 15000 - 600
m = 10000(1 + 10/100) = 10000 + 1000
Total SP - Total CP = r + m - (15000 + 10000) = -600 + 1000 = Rs. 400
As this is positive, an overall profit of Rs. 400 was made.
A
Answer: (A) s. 400
Question: A survey of n people in the town of Eros found that 50% of them preferred Brand A. Another survey of 110 people in the town of Angie found that 60% preferred Brand A. In total, 55% of all the people surveyed together preferred Brand A. What is the total number of people surveyed?
(A) 50
(B) 100
(C) 150
(D) 200
(E) 220
Rationale: It is simply a weighted average question. Since the given average of 50% and 60% is 55% (right in the middle), it means the number of people surveyed in Eros (n) is same as the number of people surveyed in Angie.
So n = 110
Total = 110 + 110 = 220
Answer (E)
Answer: (E) 220
Question: There are A girls in a class, and their average height is M inches. In the same class, there are Y boys with an average height of N inches. What is the average height of all the students in the class?
(A) (X + Y)/(M + N)
(B) (M + N)/(X + Y)
(C) (XM + YN)/(M + N)
(D) (AM + YN)/(A + Y)
(E) MN/XY
Rationale: Total height = AM+YN
No of students = X+Y
Avg = (AM+YN)/(A+Y)
Answer is D
Answer: (D) (AM + YN)/(A + Y)
Question: Sum of the squares of three numbers is 179 and the sum of their products taken two at a time is 131. Find the sum?
(A) 20
(B) 22
(C) 26
(D) 28
(E) 29
Rationale: (a + b + c)2 = a2 + b2 + c2 + 2(ab +bc + ca) = 179 + 2* 131
a + b + c = √441 = 21
B
Answer: (B) 22
Question: There are 6 ladies and 5 gents to dine at a round table. In how many ways can they seat themselves so that no two gents are seated together.
(A) 85400
(B) 86400
(C) 36400
(D) 65600
(E) 88600
Rationale: First arrange the ladies. This is a circular arrangment. 6 Gents can be arranged in 5! ways.
Now there are 6 spaces (space in between two ladies) and we have to seat 5 gents there.
First select 5 places out of 6 available places. can be done in 6C5 ways. Then arrange 5 gents there in 5! ways.
Total: 5!*6C5*5! = 86400
B is the answer.
Answer: (B) 86400
Question: In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes?
(A) 3
(B) 9
(C) 3^5
(D) 5^3
(E) 2^4
Rationale: The first letter can be posted in any of the 3 post boxes. Therefore, we have 3 possibilities.
Similarly, the second, the third, the fourth and the fifth letter can each be posted in any of the 3 post boxes.
Each of the 5 letters has 3 possibilities because we can post any number of letters in all of the boxes.
Therefore, the total number of ways the 5 letters can be posted in 3 boxes is
3 * 3 * 3 * 3 * 3 = 3^5
Ans:C
Answer: (C) 3^5
Question: In a class, there are 15 boys who are over 160 cm tall. If these constitute three-fourths of the boys and the total number of boys is two-thirds of the total number of students in the class, what is the number of girls in the class ?
(A) 10
(B) 12
(C) 18
(D) 24
(E) 26
Rationale: Explanation:
Let the number of boys be x. Then, (3/4)x = 15 or x = 15 x(4/3) = 20.
If total number of students is y, then (2/3) y = 20 or y = 20 x (3/2) = 30.
Therefore Number of girls in the class = (30 - 20) = 10.
Answer: A
Answer: (A) 10
Question: The average earning of amechanic for the first 4days of a week is Rs.60 and for the last four days is Rs.22. If heearns Rs.20 on the fourth day, his average earning forthe whole week is?
(A) s. 10
(B) s. 44
(C) s. 30
(D) s. 40
(E) s. 50
Rationale: Total earning for the week = Sum of earning of first four days + Sum of earning of last four days - Earning of 4th day
= 4 x 60 + 4 x 22 -20
= Rs. 308
âˆ´ Average earning = 308/ 7
= Rs. 44
B
Answer: (B) s. 44
Question: What sum of money will produce Rs.70 as simple interest in 4 years at 3 1/2 percent?
(A) Rs.525
(B) Rs.500
(C) Rs.550
(D) Rs.555
(E) Rs.545
Rationale: 70 = (P*4*7/2)/100
P = 500
Answer:B
Answer: (B) Rs.500
Question: Roberto has three children: two girls and a boy. All were born on the same date in different years. The sum of the ages of the two girls today is smaller than the age of the boy today, but a year from now the sum of the ages of the girls will equal the age of the boy. Two years from today, the difference between the age of the boy and the combined ages of the girls will be
(A) 1
(B) 2
(C) 3
(D) –2
(E) –1
Rationale: Approach I (Plugin's)
Girl I ............... Girl II ................ Boy
1 ....................... 1 ........................ 3 (Assume the current ages)
1 + 1 < 3 .......... (Satisfies the given condition)
1 Year later there ages are
2 ....................... 2 ......................... 4
2 + 2 = 4 ............ (Satisfies the given condition)
After 2 years there ages are
3 ....................... 3 ............................ 5
Difference = 5 - (3+3) = 5 - 6 = -1
Answer = E
Answer: (E) –1
Question: A man can row 9 kmph in still water. When the river is running at 3.1 kmph, it takes him 1 hour to row to a place and black. What is the total distance traveled by the man?
(A) 2.21
(B) 2.48
(C) 9.24
(D) 7.29
(E) 7.94
Rationale: M = 9
S = 3.1
DS = 12.1
US = 5.9
x/12.1 + x/5.9 = 1
x = 3.97
D = 3.97 * 2 = 7.94
Answer: E
Answer: (E) 7.94
Question: A particular library has 75 books in a special collection, all of which were in the library at the beginning of the month. These book are occasionally loaned out through an inter-library program. If, by the end of the month, 80 percent of books that were loaned out are returned and there are 69 books in the special collection at that time, how many books of the special collection were loaned out during that month?
(A) 20
(B) 30
(C) 35
(D) 40
(E) 55
Rationale: there are 6 books less(75-69) which represents 20% of the loaned books (100-80)
So total loaned out books = 30
Answer B
Answer: (B) 30
Question: Let y = 2m + x^2 and m = 5x + 5. If m^2 =100, then which of the following can be a value of 2y + 3m?
(A) 83
(B) 63
(C) 93
(D) 73
(E) 53
Rationale: m^2 = 100 => m=10
10= 5x+5 => x=1
y= 2*10+1^2 = 21
2y+3m = 2*21+3*10 = 93
Answer : C
Answer: (C) 93
Question: In a simultaneous throw of 2 dice what is the probability of getting a total of 3?
(A) 1/4
(B) 1/18
(C) 2/11
(D) 3/13
(E) 1/5
Rationale: n(S) = 6*6 = 36
E = {(1,2),(2,1)}
Probability = 2/36 = 1/18
Answer is B
Answer: (B) 1/18
Question: Working alone at its constant rate, machine A produces x boxes in 10 minutes and working alone at its constant rate, machine B produces x boxes in 5 minutes. How many minutes does it take machines A and B, working simultaneously at their respective constant rates, to produce 3x boxes?
(A) 3 minutes
(B) 4 minutes
(C) 10 minutes
(D) 6 minutes
(E) 12 minutes
Rationale: Rate = Work / Time
Given Rate of Machine A = X / 10 min
Machine B Produces 2x boxes in 5 min hence , Machine B produces 4x boxes in 10 min .
Rate of Machine B = 2x / 10
we need tofind the combined time that machines A and B, working simultaneouslytakeat their respective constant rates
let's first find the combined Rate of Machine A and B
Rate of Machine A = X / 10 min + Rate of Machine B = 2x / 10 = 3X/10
Now combine Time = combine work needs to be done / Combine Rate = 3x/3x * 10 = 10 Min
Ans: C
Answer: (C) 10 minutes
Question: If y exceeds x by 20%, then x is less than y by?
(A) 16 2/9%
(B) 16 3/3%
(C) 16 2/3%
(D) 12 2/3%
(E) 16 1/3%
Rationale: X=100	y=120
120------20
100-------? => 16 2/3%
Answer:C
Answer: (C) 16 2/3%
Question: The respective ages of father and his son are 41 and 16 years. In how many years will the father be twice as old as his son ?
(A) 19 years
(B) 9 years
(C) 10 years
(D) 15 years
(E) 12 years
Rationale: Suppose x years later the father will be twice as old as his son.
x + 41 = 2 (x + 16)
x = 41 - 32 = 9 years
answer : B
Answer: (B) 9 years
Question: If 9 engines consume 24 metric tonnes of coal, when each is working 8 hours day, bow much coal will be required for 8 engines, each running 13hours a day, it being given that 3 engines of former type consume as much as 4 engines of latter type?
(A) 24 metric tonnes
(B) 36 metric tonnes
(C) 14 metric tonnes
(D) 43 metric tonnes
(E) 26 metric tonnes
Rationale: Let 3 engines of former type consume 1 unit in 1 hour.
Then, 4 engines of latter type consume 1 unit in 1 hour.
Therefore 1 engine of former type consumes(1/3) unit in 1 hour.
1 engine of latter type consumes(1/4) unit in 1 hour.
Let the required consumption of coal be x units.
Less engines, Less coal consumed (direct proportion)
More working hours, More coal consumed (direct proportion)
Less rate of consumption, Less coal consumed(direct prportion)
Number of engines 9: 8
Working hours	8 : 13 } :: 24 : x
Rate of consumption (1/3):(1/4)
[ 9 x 8 x (1/3) x x) = (8 x 13 x (1/4) x 24 )
24x = 624
x = 26.
Hence, the required consumption of coal = 26 metric tonnes.
Answer is E.
Answer: (E) 26 metric tonnes
Question: The probability of having a girl is identical to the probability of having a boy. In a family with six children, what is the probability that all the children are of the same gender?
(A) 1/32
(B) 1/9
(C) 1/16
(D) 1/25
(E) 1/2
Rationale: There are two cases. All boys and all girls
case 1. all boys.
1/2*1/2*1/2*1/2*1/2*1/2= 1/64
case 2 .all girls.
1/2*1/2*1/2*1/2*1/2*1/2=1/64
We will add the cases to get 1/32
Answer: (A) 1/32
Question: Vijay bought 160 shirts at the rate of Rs. 225 per shirt. The transport expenditure was Rs. 1400. He paid an octroi at the rate of Rs. 1.75 per shirt and labour charges were Rs. 320. What should be the selling price of one shirt, if he wants a profit of 20%?
(A) Rs. 258
(B) Rs. 270
(C) Rs. 277.5
(D) Rs. 285
(E) None of these
Rationale: Total CP per shirt = 225 + 1400/160 + 1.75 + 320/160 = Rs. 237.5
SP = CP[(100 + profit%)/100]
= 237.5 * [(100 + 20)/100] = Rs. 285.
ANSWER:D
Answer: (D) Rs. 285
Question: 3897 x 999 = ?
(A) 3883203
(B) 3893103
(C) 3639403
(D) 3791203
(E) None of them
Rationale: =3897 x 999
=3897 x (1000 - 1)
=3897 x 1000 - 3897 x 1
=3897000 - 3897
=3893103
Answer is B
Answer: (B) 3893103
Question: A can complete a task in 12 days. B can complete the task in 18 days. If A and B work on this same task in alternate days starting with A, in how many days do they finish the entire task?
(A) 10.8 days
(B) 14.33 days
(C) 11 days
(D) 8.4 days
(E) 9.6 days
Rationale: DETAILED SOLUTION
A can complete a task in 12 days and B can complete the task in 18 days. So,
In one day, A can do 1/12, in one day B can do 1/18.
In two days, they can complete 1/12 + 1/18 = which is 5/36.
In 4 days, they can finish 10/36, and so on.
We can have 7 such sets of 2days each.
So, at the end of 14 days, they would have done 35/36 of task.
On the 15th day A would begin work with 1/36of the task to finish. He can finish 1/12 in a day. So, he would take one-third of a day. So, they can finish the whole task in 14.33 days.
CORRECT ANSWER B
Answer: (B) 14.33 days
Question: How many figures are required for numbering the pages of a book containing 1000 pages?
(A) 6489
(B) 3892
(C) 2893
(D) 2883
(E) 2983
Rationale: 1 to 9 = 9 * 1 = 9
10 to 99 = 90 * 2 = 180
100 to 999 = 900 * 3 =2700
1000 = 4
-----------
2893
ANSWER:C
Answer: (C) 2893
Question: The true discount on Rs. 1760 due after a certain time at 12% per annum is Rs. 160. The time after which it is due is:
(A) 4
(B) 6
(C) 8
(D) 10
(E) 12
Rationale: Sol.
P.W. = Rs. (1760 - 160) = Rs. 1600
∴ S.I. on Rs.1600 at 12% is Rs. 160.
∴ Time = [100 * 160 / 1600 * 12] = 5/6 years = [5/6 * 12] months = 10 months.
Answer D
Answer: (D) 10
Question: The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 8, respectively. If the sixth car through the production line has either 3, 7, or 12 defects, for which of theses values does the mean number of defects per car for the first six cars equal the median?
I. 3
II. 7
III. 12
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) I, II, and III
Rationale: C
for 7 we will have mean as 7.16(approx) and median as 7 but for 3 we will have mean and median = 6.5 and for 12 we will have median and mean =8
Answer: (C) III only
Question: Tough and Tricky questions: Exponents.
If 5^(x+1)*4^(y-1) = 25^x*64^y, then x + y =
(A) 3/2
(B) -1/2
(C) -1
(D) 1/2
(E) 2/3
Rationale: Here is my solution.
5^(x+1)*4^(y-1) = 25^x*64^y
Here RHS 25^x*64^y= 5^(2x)*4^(3y)
Equating powers on both sides-->
x+1=2x , thus x=1 and
2y-1=3y giving y=-1/2
So, x+y=1/2
Option: D
Answer: (D) 1/2
Question: A cave is going to explode. A family is trapped inside that cave. The only way to get outside is a narrow tunnel. But the tunnel is weak and thus only two people can go at one time. Also, everybody is afraid of going alone.
The father takes one minute to go through till the end, the mother takes two, the son takes four and the daughter takes five. The cave and tunnel will explode in twelve minutes. Can they make it to the other end in that time frame?
PS: Keep in mind that when two people are moving, the total time taken will be as per the slower one.
(A) 10 minutes
(B) 12 minutes
(C) 18 minutes
(D) 19 minutes
(E) 16 minutes
Rationale: B
12 minutes
Explanation:
Yes, surely they can all make it out of the cave.
First, father and mother will go. Time taken = 2 minutes.
Father comes back. Time taken = 2 + 1 = 3 minutes.
Both children go. Time taken = 3 + 5 = 8 minutes.
Mother comes back. Time taken = 8 + 2 = 10 minutes.
Mother and Father go. Time taken = 10 + 2 = 12 minutes.
Thus in 12 minutes, all of them will be out of the cave through the tunnel.
Answer: (B) 12 minutes
Question: 11 + 12 + 13 + ... 51 + 52 + 53 = ?
(A) 1361
(B) 1376
(C) 1363
(D) 1364
(E) 1365
Rationale: Sum = 11 + 12 + 13 + ... 51 + 52 + 53
Sum of n consecutive positive integers starting from 1 is given as n(n+1)/2
Sum of first 53 positive integers = 53*54/2
Sum of first 10 positive integers = 11*10/2
Sum = 11 + 12 + 13 + ... 51 + 52 + 53 = 53*54/2 - 11*10/2 = 1376
Answer : B
Answer: (B) 1376
Question: Z is a set of positive numbers. The median of Z is greater than the mean of Z. Which of the following Q must be true?
I. At least 50% of the numbers in Z are smaller than the median.
II. Less than 50% of the numbers in Z are greater than the median.
III. The median of Z is greater than the average of the largest and smallest numbers in Z.
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) Q=None of the above
Rationale: Condition 3 can be invalidated by: (1, 3, 4, 5)--median is equal to average of 1 and 5 and average is greater than 3.
Then, don't conditions 1 and 2 have to occur simultaneously? I reasoned this based on the word... maybe that's incorrect.
Even if not, we can rule them both out. So, E.
Answer: (E) Q=None of the above
Question: A man can row downstream at 20 kmph and upstream at 12 kmph. Find the speed of the man in still water and the speed of stream respectively?
(A) 13, 3
(B) 12, 6
(C) 16, 4
(D) 14, 4
(E) None of these
Rationale: Explanation:
Let the speed of the man in still water and speed of stream be x kmph and y kmph respectively.
Given x + y = 20--- (1)
and x - y = 12 --- (2)
From (1) & (2) 2x = 32 => x = 16, y = 4.
Answer: Option C
Answer: (C) 16, 4
Question: A man said to his son, "I was two-third of your present age when you were born". If the present age of the man is 48 years, find the present age of the son?
(A) 68.8
(B) 28.7
(C) 28.1
(D) 28.8
(E) 28.2
Rationale: Present age of the son be P, he was born P years ago.
The age of the man was: (48 - P).
His age when the son was born should be equal to 2/3 of P.
(48 - P) = 2/3 P
5P = 144 => P = 28.8
Answer:D
Answer: (D) 28.8
Question: If a certain number is divided by 2, the quotient, dividend, and divisor, added together, will amount to 62. What is the number?
(A) 18
(B) 28
(C) 40
(D) 38
(E) 59
Rationale: Let x = the number sought.
Then x/2 + x + 2 = 62.
x = 40.
C
Answer: (C) 40
Question: A zinc-copper alloy with 2 kg of pure zinc would contain 90% of zinc. The alloy with 1 kg of 90% zinc alloy contains 84% of zinc. Calculate the weight and percentage of zinc in the zinc-copper alloy.
(A) 2.4 kg or 80%
(B) 1.4 kg or 88%
(C) 1.9 kg or 94.2%
(D) 7.4 kg or 18%
(E) 1.4 kg or 60%
Rationale: Right Answer Explanation:
x : 2
y : 100
90
10 : 90 – y
(90-y)/2 = 10/x = ………..(1)
In the second case,
x 1
y 90
84
6 84-y
Therefore (84-y)/1 = 6/x ………..(2)
Dividing (1) and (2)
=
180 – 3y =840 – 10y
7y = 660
y = 94.2
x = 20/(90-80) = 20/10 = 2 kg
Quantity of zinc in x = 94.2% of x
= 0.942 2
=1.9 kg
Weight and percentage of zinc is 1.9 kg or 94.2%
C
Answer: (C) 1.9 kg or 94.2%
Question: When viewed through a certain microscope objects appear 120 times larger. If a red blood cell appears to have a radius of 49.2 centimeters, what is the actual diameter, in millimeters of the cell? (1 millimeter = 0.1 centimeter)
(A) 0.82
(B) 4.1
(C) 8.2
(D) 11.8
(E) 82
Rationale: 0.1 Centimetre = 1 millimetre
1 cm = 10 mm
49.2 cm = 49.2 * 10 mm = 492 mm
Objects are magnified by factor of 120.
Original radius = 492/120= ~4.1
Diameter = 4.1 * 2 = 8.2
Ans C.
Answer: (C) 8.2
Question: A man can hit a target once in 6 shots. If he fires 4 shots in succession, how many ways that he will hit his target?
(A) 71/234
(B) 81/133
(C) 81/343
(D) 27/56
(E) 91/324
Rationale: The man will hit the target if he hits it once or twice or thrice or all four times in the four shots that he takes.
So, the only possibility when the man will not hit the target is when he fails to hit the target in even one of the four shots that he takes.
The event of not hitting the target even once is the complement of the event of hitting the target at least once.
The probability that he will not hit the target in any given shot = 1 - 1/6 = 5/6
Therefore, the probability that he will not hit the target in any of the four shots = 5/6^4 =27/56
D
Answer: (D) 27/56
Question: The current of a stream at 1 kmph. A motor boat goes 35 km upstream and back to the starting point in 12 hours. The speed of the motor boat in still water is?
(A) 6
(B) 9
(C) 8
(D) 3
(E) 1
Rationale: S = 1
M = x
DS = x + 1
US = x - 1
35/(x + 1) + 35/(x - 1) = 12
x = 6
Answer: A
Answer: (A) 6
Question: P cando a work in 15 days and Q cando the same work in 20 days. If they can work together for 4 days, what is the fraction of work left?
(A) 2/13
(B) 3/17
(C) 8/15
(D) 9/21
(E) 9/22
Rationale: Amount of work P can do in 1 day = 1/15
Amount of work Q can do in 1 day = 1/20
Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60
Amount of work P and Q can together do in 4 days = 4 × (7/60) = 7/15
Fraction of work left = 1 – 7/15= 8/15
C
Answer: (C) 8/15
Question: In an examination a pupil's average marks were 63 per paper. If he had obtained 18 more marks for his Geography paper and 2 more marks for his History paper, his average per paper would have been 65. How many papers were there in the examination?
(A) 8
(B) 9
(C) 10
(D) 11
(E) 12
Rationale: Let the number of papers be x. Then,
63x + 18 + 2 = 65x
= 2x = 20
= x = 10.
ANSWER:C
Answer: (C) 10