mathewhe/aqua_rat huggingface dataset:
Question: A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and
second half at the rate of 24 km/hr. Find the total journey in km
(A) 200 Km
(B) 222 Km
(C) 224 Km
(D) 248 Km
(E) None of these
Rationale: Explanation:
Let time taken to travel the first half = x hr
Then time taken to travel the second half = (10 - x) hr
Distance covered in the the first half = 21x [because, distance = time*speed]
Distance covered in the the second half = 24(10 - x)
Distance covered in the the first half = Distance covered in the the second half
So,
21x = 24(10 - x)
=> 45x = 240
=> x = 16/3
Total Distance = 2*21(16/3) = 224 Km [multiplied by 2 as 21x was distance of half way]
Answer: C
Answer: (C) 224 Km
Question: A semicircle has a radius of 9. What is the approximate perimeter of the semicircle?
(A) 16
(B) 25
(C) 33
(D) 46
(E) 58
Rationale: The perimeter of a circle is 2*pi*r.
The perimeter of a semicircle is 2*pi*r/2 + 2r = pi*r + 2r
The perimeter is pi*9 + 2*9 which is about 46.
The answer is D.
Answer: (D) 46
Question: The probability of two events A and B are 0.20 and 0.40 respectively. The probability that both A and B occur is 0.15. The probability that neither A nor B occur is _________
(A) 0.45
(B) 0.4
(C) 0.5
(D) 0.55
(E) 0.6
Rationale: we are apply that formula..............
P(AorB)=P(A)+P(B)-P(A AND B)
=.20+.40-.15
=.45
but the probability of neither A nor B=1-.45
=0.55
ANSWER:D
Answer: (D) 0.55
Question: A can work four times as fast as B. Together they can finish a job in 12 days. In how many days can A complete the job alone?
(A) 14
(B) 15
(C) 16
(D) 17
(E) 18
Rationale: B's rate is x and A's rate is 4x.
x + 4x = 1/12
x = 1/60
A's rate is 4x = 1/15
A can complete the job in 15 days.
The answer is B.
Answer: (B) 15
Question: 100 identical coins each with probability 'pp' showing up heads & tossed.
(A) 13/101
(B) 11/101
(C) 51/101
(D) 53/102
(E) 54/101
Rationale: Let aa be the number of coins showing heads
Then, P(A=50)=P(A=51)P(A=50)=P(A=51)
⇒100C50×P50×(1−P)50⇒100C51×P51×(1−P)49⇒100C50×P51×(1−P)50=100C51×P⇒51(1âˆ′P)=50P⇒P=51/101
C
Answer: (C) 51/101
Question: If a#b = ab – b + b^2, then 3#4 =
(A) 2
(B) 8
(C) 15
(D) 24
(E) 35
Rationale: Solution -
Simply substitute 3 and 4 in equation in the place of a and b respectively.
3#4 = 3*4 - 4 + 4^2 = 12 - 4 + 16 = 24. ANS D
Answer: (D) 24
Question: How many different four letter words can be formed (the words need not be meaningful using the letters of the word "MEDITERRANEAN" such that the first letter is E and the last letter is R?
(A) 59
(B) 77
(C) 36
(D) 99
(E) 17
Rationale: Explanation:
The first letter is E and the last one is R.
Therefore, one has to find two more letters from the remaining 11 letters.
Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.
The second and third positions can either have two different letters or have both the letters to be the same.
Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.
Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.
Total number of possibilities = 56 + 3 = 59
Answer: A) 59
Answer: (A) 59
Question: The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
(A) 101
(B) 107
(C) 111
(D) 185
(E) 186
Rationale: Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
Answer: Option C
Answer: (C) 111
Question: Find the value of 85P3 .
(A) 565350
(B) 595650
(C) 535950
(D) 565350
(E) 575350
Rationale: 85P3 = 85!/(85-3)!
= 85!/82!
= 85*84*83*82!/82!
= 85*84*83
= 595650
ANSWER:B
Answer: (B) 595650
Question: A pet store regularly sells pet food at a discount of 10 percent to 30 percent from the manufacturer’s suggested retail price. If during a sale, the store discounts an additional 20 percent from the discount price, what would be the lowest possible price of a container of pet food that had a manufacturer’s suggested retail price o f $ 30.00?
(A) $ 10.00
(B) $ 11.20
(C) $ 14.40
(D) $ 16.80
(E) $ 18.00
Rationale: For retail price = $30
First maximum discounted price = 30 - 30% of 30 = 30-9 = 21
Price after additional discount of 20% = 21 - 20% of 21 = 21 - 4.2 = 16.8
Answer: Option D
Answer: (D) $ 16.80
Question: About the number of pairs which have 20 as their HCF and 136 as their LCM, the conclusion can be
(A) only one such pair exists
(B) only two such pairs exist
(C) many such pairs exist
(D) yes such pair exists
(E) paid does not exists
Rationale: Explanation:
HCF is always a factor of LCM. ie., HCF always divides LCM perfectly.
Correct Option: E
Answer: (E) paid does not exists
Question: If two different solutions of alcohol with a respective proportion of water to alcohol of 3:1 and 2:3 were combined, what is the concentration of alcohol in the new solution if the original solutions were mixed in equal amounts?
(A) 30.0%
(B) 36.6%
(C) 42.5%
(D) 44.4%
(E) 60.0%
Rationale: Alcohol: Water
First mixture -- 3:1 -- 15:5
Second mixture -- 2:3 -- 8:12
Equal amounts of the mixture were combined, hence the final mixture will be in the ratio -- (15+8): (5+12) -- 23:17
% of alcohol -- 17/(23+17) * 100
= (17/40) * 100
= 42.5
ANSWER:C
Answer: (C) 42.5%
Question: How many prime numbers are between 25/8 and 213/6?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
Rationale: 25/8 = 4-
213/6 = 36-
Prime numbers between 4 and 36 are 5, 7, 11, 13, 17, 19, 23, 29, and 31
- sign signifies that the number is marginally less .
Answer D
Answer: (D) 9
Question: Rs.20 is the true discount on Rs. 260 due after a certain time. What will be the true discount on the same sum due after half of the former time, the rate of interest being the same?
(A) 10.4
(B) 11
(C) 14.8
(D) 15.4
(E) none
Rationale: Sol.
S.I. on Rs. (260 - 20) for a given time = Rs. 20.
S.I. on Rs. 240 for half the time = Rs. 10.
T.D. on Rs. 250 = Rs. 10.
∴ T.D. on Rs. 260 = Rs. [10/250 * 260] = Rs. 10.40
Answer A
Answer: (A) 10.4
Question: If m^(-1) = -1/6 then m^(-2) is equal to
(A) -9
(B) -3
(C) -1/9
(D) 1/9
(E) 1/36
Rationale: You made a mistake here.
You're right till
m^-2 = 1/m^2 and m^-1 = 1/m^1 = 1/m
given m^-1 = -(1/6) so, 1/m = -(1/6) solving this, m = -6
Now, m^-2 = 1/m^2 = 1/(-6)^2 = 1/36 ANS E
Answer: (E) 1/36
Question: If the cost price of 165 chocolates is equal to the selling price of 150 chocolates, the gain percent is
(A) 8
(B) 9
(C) 10
(D) 11
(E) 12
Rationale: Let C.P. of each chocolate be Re. 1.
Then, C.P. of 150 chocolates = Rs. 150; S.P. of 150 chocolates = Rs. 165.
Gain % = 15/150 * 100 = 10%
Answer:C
Answer: (C) 10
Question: What number comes next?
284, 693, 712, 846, 937, ?
(A) 113
(B) 231
(C) 245
(D) 128
(E) 123
Rationale: D
128
The numbers 28469371 are being repeated in the same sequence.
Answer: (D) 128
Question: In a farm, 2000 seeds were planted wheat and 3000 rice seeds were planted. If 57 percent of the seeds in plot I germinated and 42 percent of the seeds in plot II germinated, what percent of the total number of planted seeds germinated?
(A)  45.5%
(B)  46.5%
(C)  48.0%
(D)  49.5%
(E)  51.0%
Rationale: Plot I: 57% of 2000 wheat seeds germinated, hence 0.57*2000 = 1140 seeds germinated.
Plot II: 42% of 3000 rice seeds germinated, hence 0.42*3000 = 1260 seeds germinated.
Thus (germinated)/(total) = (1140 + 1260)/(2000 + 3000) = 2400/5000 = 48% seeds germinated.
Answer: C.
Answer: (C)  48.0%
Question: Of the students at a certain high school, 90 percent took a course in algebra or geometry and 30 percent took courses in both. If the percent of students who took a course in algebra was 2 times the percent of students who took a course in geometry, what percent of the students took a course in algebra but not geometry?
(A) 50%
(B) 40%
(C) 45%
(D) 35%
(E) 30%
Rationale: You can also just use the formula
n(A or B) = n(A) + n(B) - n(A and B)
90 = 2X + X - 30
X = 40
2X = 80
This is the number of people who took algebra including those who took both. To get the number of people who took only algebra, 80 - 30 = 50%
Answer (A)
Answer: (A) 50%
Question: If x and y are positive real numbers such that x + y = 1, which of the following could be the value of 100x + 400y?
I. 80
II. 250
III. 399
(A) II only
(B) III only
(C) I and II
(D) I and III
(E) II and III
Rationale: We know that x and y are both between 0 and 1.
The minimum limit of 100x + 400y is 100, when x is close to 1 and y is close to 0.
The maximum limit of 100x + 400y is 400, when x is close to 0 and y is close to 1.
The expression can take on any value between 100 and 400.
The answer is E.
Answer: (E) II and III
Question: David obtained 72, 60, 35, 62 and 84 marks (out of 100) in English, Mathematics,Physics, Chemistry and Biology What are his average marks ?
(A) 75
(B) 62.6
(C) 87
(D) 165
(E) 11
Rationale: Explanation:
Average = (72 +60+ 35+ 62+84 )/ 5 = 313/5 = 62.6.
Answer:B
Answer: (B) 62.6
Question: If a = (3)b and c = 5b, then which of the following represents the average (arithmetic mean) of a, b, and c, in terms of b ?
(A) b/3
(B) b/4
(C) 2b
(D) 3b
(E) 5b
Rationale: a = (3)b and c =5b.
The mean = (a + b + c)/3 = (3b + b + 5b)/3 = 3b.
Answer is D
Answer: (D) 3b
Question: By selling an umbrella for Rs. 300, a shop keeper gains 20%. During a clearance sale, the shopkeeper allows a discount of 10% on the marked price. His gain percent during the sale is?
(A) 5
(B) 6
(C) 8
(D) 9
(E) 1
Rationale: Marked price = Rs. 300
C.P. = 100/120 * 300 = Rs. 250
Sale price = 90% of Rs. 300 = Rs. 270
Required gain % = 20/250 * 100 = 8%.
Answer:C
Answer: (C) 8
Question: In a workshop there are 3 kinds of beds, 3 kinds of closets, 2 kinds of shelves and 7 kinds of chairs. In how many ways can a person decorate his room if he wants to buy in the workshop one shelf, one bed and one of the following: a chair or a closet?
(A) 168.
(B) 80.
(C) 60.
(D) 48.
(E) 16.
Rationale: ways to select one shelf = 2
ways to select one bed = 3
ways to select one of the following: a chair or a closet = 7+3 = 10
ways to select all of them together = 2*3*10 = 60
Hence, C is the answer.
Answer: (C) 60.
Question: The simple form of the ratio 4/3:3/7 is?
(A) 7:3
(B) 10:3
(C) 15:3
(D) 25:3
(E) 30:3
Rationale: 4/3 : 3/7 = 21:9 = 7:3
ANSWER:A
Answer: (A) 7:3
Question: 16 chess players take part in a tournament. Every player plays twice with each of his opponents. How many games are to be played?
(A) 190
(B) 200
(C) 210
(D) 220
(E) 240
Rationale: 2*16C2 = 2*120 = 240
The answer is E.
Answer: (E) 240
Question: A vessel of capacity 90 litres is fully filled with pure milk. Nine litres of milk is removed from the vessel and replaced with water. Nine litres of the solution thus formed is removed and replaced with water. Find the quantity of pure milk in the final milk solution?
(A) 72.9
(B) 78.5
(C) 25.4
(D) 31.5
(E) 70.5
Rationale: Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively.
Quantity of milk finally in the vessel is then given by [(T - y)/T]n * T
For the given problem, T = 90, y = 9 and n = 2.
Hence, quantity of milk finally in the vessel
= [(90 - 9)/90]2 (90) = 72.9 litres.
ANSWER A
Answer: (A) 72.9
Question: The calendar for the year 2001 is same for which of the following year?
(A) 2005
(B) 2007
(C) 2011
(D) 2006
(E) 2009
Rationale: SOLUTION :
The total number of odd days from 2001 onwards should be zero.
Now, as an ordinary year adds 1 odd day and a leap year adds 2 odd days, we have:
2001, 2002, 2003, 2005, 2006 – 1 odd day each
2004 – 2 odd days
Hence, at the end of 2006 total number of odd days = 7 or 0
Therefore, the calendar for the year 2001 is repeated in the year 2007.
ANSWER IS B
Answer: (B) 2007
Question: A seven-digit phone number is divisible by 4. After adding the two digits of the area code the number is not divisible by 4, which of the following cannot be the area code?
(A) 07.
(B) 06.
(C) 05.
(D) 04.
(E) 02.
Rationale: Given that 7 digit number is divisible by 4
we know any number will not be divisible by 4 if the sum of all digits of the number is not the multiple of 4.
Hence looking at the option we can say that the number that should be added to the seven digit number must be 04 as adding it will make it divisible by 4... hence answer is D
Answer: (D) 04.
Question: How many trucks are there if each truck carrying 70 packages and total of 490 packages?
(A) A)7
(B) B)6
(C) C)9
(D) D)11
(E) E)None of the above
Rationale: Sol.
Total packages 490
Each truck carries 70 packages
= 490 / 70
= 7
Answer : A
Answer: (A) A)7
Question: If R is a positive integer, and R/15 = 8.2, what is the remainder when R is divided by 15?
(A) A.1
(B) B.2
(C) C.3
(D) D.4
(E) E.8
Rationale: If R were a multiple of 15, the quotient R/15 would be an integer. The fact that it's a decimal tells us that 15 goes into R eight whole times and some decimal part of a time. This decimal part, 0.2 is the remainder divided by the divisor. Let B be the remainder.
B/15 = 0.2 = 1/5
B = (15)*(1/5) = 3
Answer =(C)
Answer: (C) C.3
Question: In shop contains 500 articles each of a cost of $10. He sold them at a price of $11. If the shop keeper has a order of 200 articles, how much profit he will get?
(A) A)2100
(B) B)2000
(C) C)1500
(D) D)200
(E) E)600
Rationale: Total money after selling articles = 200*11 = 2200
cost price = 200*10 = 2000
profit = 200
correct option is D
Answer: (D) D)200
Question: A man has some hens and cows. If the number of heads be 48 and the number of feet equals 144, then the number of hens will be:
(A) 22
(B) 23
(C) 24
(D) 26
(E) 28
Rationale: let hens be x and cows be y
now,
feet: x*2+y*4=144
heads: x*1+y*1=48
implies, 2x+4y=144 and x+y=48
solving these two equations, we get x=24 and y=24
Therefore, hens are 26.
ANSWER:C
Answer: (C) 24
Question: There r 4 prime numbers in ascending order. The multiplication of first 3 is 385 and that of last 3is 1001. what is the last number is
(A) 10
(B) 12
(C) 11
(D) 13
(E) 15
Rationale: Explanation:
abcbcd=3851001=>ad=513
abcbcd=3851001=>ad=513
So d = 13
D
Answer: (D) 13
Question: Which of the numbers below is the odd one out
1,3,7,13,24,21,31.
(A) 3
(B) 7
(C) 21
(D) 24
(E) 31
Rationale: 24 is the odd man.. diff between sucessive no.s are multiples of 2
ANSWER:D
Answer: (D) 24
Question: Amar takes as much time in running 18 meters as a car takes in covering 48 meters. What will be the distance covered by Amar during the time the car covers 1.6 km?
(A) 277
(B) 289
(C) 269
(D) 600
(E) 122
Rationale: Distance covered by Amar = 18/4.8 (1.6km)
= 3/8(1600)
= 600 m
Answer: D
Answer: (D) 600
Question: What will be 60 percent of a number whose 200 percent is 90 ?
(A) 144
(B) 72
(C) 27
(D) None of these
(E) Cannot be determined
Rationale: Answer
∵ 200% of N = 90
⇒ N = (90 x 100) / 200 = 45
∴ 60% of N = (60 x 45) / 100 = 27
Correct Option: C
Answer: (C) 27
Question: Find the smallest number in a GP whose sum is 38 and product 1728
(A) 12
(B) 20
(C) 8
(D) All of these
(E) none of these
Rationale: In GP :- a/r+a+ar....=38
a(1+r+r^2)=38r......eq1
Product a^3=1728
a=12
now put this value in eq1
12(1+r+r^2)=38r
r=2/3 and 3/2..so sallest number is 8
ANSWER:C
Answer: (C) 8
Question: Of the teams competing in the world archery championships, Forty percent are from Europe. Half as many are from the United States and one twentieth are from Africa. What fraction of teams are from neither Europe, the US or Africa.
(A) 13/20
(B) 15/20
(C) 13/15
(D) 17/20
(E) 7/17
Rationale: Assume 100 total people.
40% of 100 = 40 --->Europeans
1/2 of 40 ---> 20 Americans
1/20 of 100 ---> 5 Africans
Now you would add them all up, and minus it by the total (100). 100 - 65= 35. So, 35 people out of 100 are not EU, US or Africans. 35/100 is also equal to 7/20.
So therefore, the answer is E.
Answer: (E) 7/17
Question: A jar full of whisky contains 30% alcohol. A part of this whisky is replaced by another containg 19% alcohol and now the percentage of alcohol was found to be 26%. What quantity of whisky is replaced ?
(A) 1/3
(B) 2/3
(C) 2/5
(D) 3/5
(E) 4/5
Rationale: Let us assume the total original amount of whiskey = 10 ml ---> 4 ml alcohol and 6 ml non-alcohol.
Let x ml be the amount removed ---> total alcohol left = 4-0.4x
New quantity of whiskey added = x ml out of which 0.19 is the alcohol.
Thus, the final quantity of alcohol = 4-0.4x+0.19x ----> (4-0.21x)/ 10 = 0.26 ---> x = 20/3 ml.
Per the question, you need to find the x ml removed as a ratio of the initial volume ---> (20/3)/10 = 1/3.
Hence, A is the correct answer.
Answer: (A) 1/3
Question: At a dinner party, 5 people are to be seated around a circular table. 2 seating arrangements are considered different only when the positions of the people are different relative to each other. what is the total number E of different possible seating arrangements for the group?
(A) 5
(B) 10
(C) 24
(D) 32
(E) 120
Rationale: Circular arrangements= (n-1)!
Hence (4!) = 24=E
Answer is C
Answer: (C) 24
Question: If a and b are both odd numbers, which of the following is an odd number?
(A) a + b
(B) a + b + 2
(C) ab+2
(D) ab + 1
(E) NOne
Rationale: Sol.
Product of two odd numbers is always odd and adding 2 gives odd.
Answer C
Answer: (C) ab+2
Question: If y is 70% greater than x, than x is what % less than y?
(A) 18.7
(B) 25.5
(C) 33.3
(D) 41.2
(E) 54.4
Rationale: y = 1.7x
x = y/1.7 = 10y/17
x is 7/17 less which is about 41.2% less than y.
The answer is D.
Answer: (D) 41.2
Question: 31 of the scientists that attended a certain workshop were Wolf Prize laureates, and 14 of these 31 were also Nobel Prize laureates. Of the scientists that attended that workshop and had not received the Wolf prize, the number of scientists that had received the Nobel Prize was 3 greater than the number of scientists that had not received the Nobel Prize. If 50 of the scientists attended that workshop, how many of them were Nobel Prize laureates?
(A) A)11
(B) B)18
(C) C)25
(D) D)29
(E) D)36
Rationale: lets solve by creating equation..
W = 31..
Total = 50..
Not W = 50 -31= 19..
Now let people who were neither be x, so out of 19 who won Nobel = x+3..
so x+x+3=19 or x=8..
so who won Nobel but not Wolf = x+3 = 11..
But people who won BOTH W and N = 13..
so total who won N = 11+14 = 25..
C
Answer: (C) C)25
Question: The difference between the simple interest received from two different sources on Rs. 1500 for 3 years is Rs. 13.50. The difference between their rates of interest is?
(A) 0.9%
(B) 0.4%
(C) 0.3%
(D) 0.2%
(E) 0.1%
Rationale: (1500 * R1 * 3)/100 - (1500 * R2 * 3)/100 = 13.50 4500(R1 - R2) = 1350
R1 - R2 = 0.3%
Answer: C
Answer: (C) 0.3%
Question: On dividing a number by 5, we get 5 as quotient and 0 as remainder. On dividing the same number by 11, what will be the remainder?
(A) 5
(B) 2
(C) 7
(D) 9
(E) 3
Rationale: Number = 5 * 5+ 0 = 25
11) 25 (2
22
--------
3
Required number = 0.
ANSWER:E
Answer: (E) 3
Question: A, B and C are partners. A receives 2/3 of profits, B and C dividing the remainder equally. A's income is increased by Rs.200 when the rate to profit rises from 5 to 7 percent. Find the Capital of B?
(A) 3999
(B) 7799
(C) 2500
(D) 2772
(E) 2912
Rationale: A:B:C = 2/3:1/6:1/6 = 4:1:1
x * 2/100 * 2/3 = 200
B capital = 15000*1/6 = 2500.Answer: C
Answer: (C) 2500
Question: What is the sum of the multiples of 4 between 19 and 130 inclusive?
(A) 2012
(B) 2044
(C) 2060
(D) 2072
(E) 2080
Rationale: the fastest way in an AP is to find the average and multiply with total integers..
between 19 and 130, the smallest multiple of 4 is 20 and largest =128..
average=(20+128)/2=148/2=74.
total numbers = (128-20)/4 + 1= 27+1=28..
sum=28*74=2072
ans D
Answer: (D) 2072
Question: The average of first five prime numbers greater than 3 is?
(A) 32.2
(B) 10.6
(C) 32.3
(D) 32.8
(E) 32.4
Rationale: 5 + 7 + 11 + 13 + 17
= 53/5
= 10.60
Answer = B
Answer: (B) 10.6
Question: Find compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annually
(A) Rs 312
(B) Rs 412
(C) Rs 512
(D) Rs 612
(E) None of these
Rationale: Explanation:
Please apply the formula
Amount=P(1+R100)nC.I. = Amount - P
Answer: D
Answer: (D) Rs 612
Question: A can contains a mixture of liquids A and B is the ratio 7:5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. How many liter E of liquid A was contained by the can initially?
(A) 10
(B) 20
(C) 21
(D) 25
(E) 27
Rationale: As A:B::7:5 ---> only option C is a multiple of 7 and hence it is a good place to start. Also A:B::7:5 means that , A = (712)*Total and B = (5/12)*Total
If A = 21 , B = 15 ---> remove 9 litres ---> you remove (7/12)*9 of A ---> A remaining = 21-(7/12)*9 = 63/4
Similarly, for B, you remove (5/12)*9 ---> B remaining = 15 - (5/12)*9 = 45/4 and then add 9 more litres of B ---> 9+45/4 = 81/4
Thus A/B (final ratio) = (45/4)/(81/4) = 7:9 , the same as the final ratio mentioned in the question.
Hence C is the correct answer.
A/B = 7/9 = (7x-(7/12)*9)/ (5x-(5/12)*9+9), where 7x and 5x are initial quantities of A and B respectively.
Thus, 7/9 = (7x-(7/12)*9)/ (5x-(5/12)*9+9) ---> giving you x=3. Thus A (original) E= 7*3 = 21.C
Answer: (C) 21
Question: Kate is 12 years old. His age is 4 times the age of Robbie. Find the age of Robbie.
(A) 1yrs
(B) 2yrs
(C) 3yrs
(D) 4yrs
(E) 5yrs
Rationale: Let Kate age, x = 12
robbie = x/4
=12/4
=3yrs
ANSWER:C
Answer: (C) 3yrs
Question: Five years ago, the average age of A, B, C and D was 45 years. With E Joining them now, the average of all the five is 51 years. The age of E is?
(A) 45
(B) 47
(C) 55
(D) 49
(E) 46
Rationale: Solution
5 years ago average age of A,B,C,D = 45 years
=> 5 years ago total age of A, B, C, D = 45 x 4 = 180 years
=> Total present age of A, B, C, D = 180 + 5x4 = 200 years
If E's present age is x years = 200+x/5= 51
x=55 years. Answer C
Answer: (C) 55
Question: At a certain high school, a student’s popularity is determined by his or her locker number. Whoever has the locker number with the greatest number of distinct prime factors is the most popular student in the school. If Dhoni ,Kamal, Gokul, and Dinesh get lockers with the numbers 50, 400, 150, and 25, respectively, who is the most popular student?
(A) Dhoni
(B) Kamal
(C) Gokul
(D) Dinesh
(E) They have equal popularity
Rationale: Distinct prime factors of 50 (Dhoni) : 2,5
Distinct prime factors of 400 (Kamal): 2 and 5.
Distinct prime factors of 150 (Gokul): 2,3, and 5
Distinct prime factors of 25 (Dinesh): 5.
So most popular student - Gokul
Answer C
Answer: (C) Gokul
Question: If 4 and 11 are the lengths of two sides of a triangular region, which of the following can be the length of the third side?
I. 5
II. 13
III. 15
(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, and III
Rationale: The length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides.
i.e let us assume the third side be c and other sides be a and b, then as per rule
c > a - b and c < a + b.
Let's take options
option 1. the three sides will be 5 , 4 and 11..
then 11 > 5-4 --Yes
11 < 4+9...NO
option 2... 13, 4 and 11.
then 13 > 11-4...Yes
13 < 11+4...Yes
Option 3..15 ,4 and 11.
15 < 11+4...No
15 > 11-4...Yes...
Only 13 is correct .
Answer: option B is correct..
Answer: (B) II only