mathewhe/aqua_rat huggingface dataset:
Question: S is a set containing 6 different positive odd primes. T is a set containing 8 different numbers, all of which are members of S. Which of the following statements CANNOT be true?
(A)  The median of S is prime.
(B)  The median of T is prime
(C)  The median of S is equal to the median of T.
(D)  The sum of the terms in S is prime.
(E)  The sum of the terms in T is prime.
Rationale: Here is my explanation. The question states:S is a set containing 9 different positive odd primes. T is a set containing 8 different numbers, all of which are members of S. Which of the following statements CANNOT be true?
(A) The median of S is prime.
Thismust be true. If there are an odd number of members of a set, then the median is a member of the set: it's the middle number, when all the numbers are ranked from smallest to biggest. Every number in S is a positive odd prime, so the median is one of them, and is prime.
(B) The median of T is prime.
Thismay or may not be true. If a set has an even number of members, the median is average of the two numbers in the middle, when ranked from smallest to biggest. The average of two odd numberscould beeven (average of 71 and 73 is 72), and hence not prime, or itcould beodd (the average of 71 and 79 is 75). For particularly well chosen odd numbers, the average can be not only odd but also prime -- for example, the average of 89 and 113 is 101, another prime number. If the two middle numbers of T were 89 and 113, the median would be 101, a prime number.
(C) The median of S is equal to the median of T.
Under most configurations for S and T, this wouldn't happen. If you weren't trying to make it happen, it would be unlikely to happen by chance. BUT, if the number dropped going from from S to T was the median of S (say, 101), and if the two middle numbers of T happen to have an average of that number that was dropped (for example, if the two numbers were 89 and 113), then the medians would be equal. In other words, the three middle numbers of S would have to be {. . ., 89, 101, 133, . . .}, and when 101 is dropped in going to T, the median of two would be the average of 89113, which happens to be 101. It's an exceptional case, but itcouldbe true.
(D) The sum of the terms in S is prime.
Thismay or may not be true. The sum of 9 odd numbermustbe an odd number. That odd numbercould beprime. For example, the sum of the first nine odd prime numbers {3, 5, 11, 13, 17, 19, 23, 29} is 127, which is prime. If you drop 3 and include the next prime, 31, the set {5, 11, 13, 17, 19, 23, 29, 31} has a sum of 155, which is clearly not prime.
(E) The sum of the terms in T is prime.
Thismust be false. The sum of eight odd numbers must be an even number. Only 2 is prime, and all other even numbers are not. Therefore, the sum of eight odd prime numbers will be an even number bigger than two, and absolutely cannot be prime.
D
Answer: (D)  The sum of the terms in S is prime.
Question: A child must place 6 different toys in 4 different bins. If any of the toys can go in any of the bins, in how many ways can the child place the toys into the bins?
(A) 6^6
(B) 6^4
(C) 4^6
(D) 4^4
(E) 6^4-4^6
Rationale: there are 4 different bins and each of the toy can go to any of these bins..
so answer 4∗4∗4∗4∗4∗4=4^6
ANSWER:C
Answer: (C) 4^6
Question: If –1 < x <0 and x ≠ 0, which of the following inequalities must be true?
I. x^3 < x
II. x^2 < |x|
III. x^4 – x^5 > x^3 – x^2
(A)  I only
(B)  II only
(C)  III only
(D)  II and III only
(E)  I, II, and III
Rationale: Answer is D. Let's look at the statements one by one
Stmt I. x^3 < x
if 0<x<1 then x^3< x but if -1<x<0 then x^3>x
So this statement is not always true
Stmt II. x^2 < |x|
Because we know that x is a number less than one but not equal to zero then x^2 will always be less than |x|.
Why? think of positive fractions (and you can think in terms of positive fractions because the inequality is in regards to |x|). Lets set x = 1/2, then x^2 = 1/4 and 1/4<1/2
So Stmt II is always true
Stmt III. x^4 – x^5 > x^3 – x^2
This one may seem tricky but lets break it down. x^4 – x^5 > x^3 – x^2 = x^4(1-x)>x^2(x-1).
Because lets concentrate on (1-x) and (x-1). We are given that -1<x<1 which means that (1-x)>0 and (x-1)<0. x^4 will always be positive and x^2 will always be positive so without doing any math we are looking at positive > negative... which is always true.
So Stmt III is always true
B
Answer: (B)  II only
Question: If the number 91876*2 is completely divisible by 8, then the smallest whole number in place of * will be?
(A) 1
(B) 2
(C) 3
(D) 4
(E) None of these
Rationale: The number 6x2 must be divisible by 8.
x=3 as 632 is divisible by 8.
Correct Option : C
Answer: (C) 3
Question: A man walking at a rate of 15km/hr crosses a bridge in 10 minutes. The length of the bridge is?
(A) 500m
(B) 1110m
(C) 950m
(D) 1000m
(E) 1300m
Rationale: Speed = 15*5/18 = 15/18 m/sec
Distance covered in 10 minutes = 15/18 *10 *60 = 500 m
Answer is A
Answer: (A) 500m
Question: which number need to add to 859622 to get a number exactly divisible by 456?
(A) 678311
(B) 546576
(C) 465767
(D) 678980
(E) 859622
Rationale: Dividend = Quotient * Divisor + Reminder
859622 / 456 gives quotient = 1885 and reminder = 62.
So, the next number divisible by 456 is 456 places infront of 456 * 1885
Which means 456 – 62 = 394 should be added to 859622.
E
Answer: (E) 859622
Question: 10 books are placed at random in a shelf. The probability that a pair of books will always be together is
(A) 1/10
(B) 9/10
(C) 1/5
(D) 3/10
(E) 1/2
Rationale: 10 books can be rearranged in 10! ways consider the two books taken as a pair then number of favourable ways of getting these two books together is 9! 2!
Required probability = 1/5
ANSWER:C
Answer: (C) 1/5
Question: A company bought for its 6 offices 2 computers of brand N and 3 computers of brand M. In how many ways could computers be distributed among the offices if each office can have only 1 computer.
(A) 60
(B) 210
(C) 256
(D) 292
(E) 312
Rationale: This problem has to do withcombinations. Here's the general idea: if you have a set of n elements, and you are going to choose r of them (r < n), then the number of combinations of size r one could choose from this total set of n is:
# of combinations = nCr = (n!)/[(r!)((n-r)!)]
where n! is the factorial symbol, which means the product of every integer from n down to 1. BTW, nCr is readn choose r.
In this problem, let's consider first the three computers of brand M. How many ways can three computer be distributed to seven offices?
# of combinations = 6C3 = 20
There are 20 different ways to distribute three computers to 7 offices. (The massive amount of cancelling that occurred there is very much typical of what happens in the nCr formula.)
One we have distributed those three M computers, we have to distribute 2 N computers to the remaining four offices. How many ways can two computer be distributed to four offices?
# of combinations = 3C2 = 3
For each of the 20 configurations of distributing the M computers, we have 3 ways of distributing the N computers to the remaining offices. Thus, the total number of configurations is 20 * 3 = 60. Answer choice =A
Answer: (A) 60
Question: A total of 30 percent of the geese included in a certain migration study were male. If some of the geese migrated during the study and 25 percent of the migrating geese were male, what was the ratio of the migration rate for the male geese to the migration rate for the female geese? [Migration rate for geese of a certain sex = (number of geese of that sex migrating) / (total number of geese of that sex)]
(A) 1/4
(B) 7/12
(C) 2/3
(D) 7/8
(E) 9/7
Rationale: Let' take the Number of geese to be 100.
Male = 30. Female = 70.
Now the second part of the Q, Let's take the Number migrated to be 20.
So we have 20 geese that migrated and out of that 25% are male i.e 25/100 *20 = 5 Geese (Males)
And now we know out of the total 20 geese , 5 are male , then 15 have to be female.
Now the ratio part , Male Geese Ratios = 5/30 = 1/6. - A
Female Geese Ratios =15/70=3/14 - B
Cross Multiply Equations A and B and you get =9/7 .
Ans E
Answer: (E) 9/7
Question: What is the next number of the following sequence
21,77,165,285,......
(A) 434
(B) 435
(C) 436
(D) 437
(E) 438
Rationale: 21=5^2-4
77=9^2-4
165=13^2-4
285=17^2-4
5,9,13,17,21(+4)
in the same way next no. will be
21^2-4=441-4=437
ANSWER:D
Answer: (D) 437
Question: A, B and C rents a pasture for Rs.870. A put in 12 horses for 8 months, B 16 horses for 9 months and 18 horses for 6 months. How much should A pay?
(A) 240
(B) 227
(C) 268
(D) 198
(E) 176
Rationale: 12*8 :16*9 = 18*6
8: 12: 9
8/29 * 870 = 240
Answer:A
Answer: (A) 240
Question: Two cubes have their volumes are in the ratio 1:27. Find the ratio of their surface area?
(A) 1:2
(B) 2:3
(C) 1:9
(D) 2:7
(E) 3:11
Rationale: Let their edges be a and b
Then, a^3 / b^3 = 1/27
(a/b)^3 = (1/3)^3
a/b = 1/3
Ratio of their surface areas = 6a^2 / 6b^2 = (a/b)^2 = 1/9 = 1:9
Answer is C
Answer: (C) 1:9
Question: If ty > 0 and yz < 0, which of the following must be negative:
(A) tyz
(B) ty(z^2)
(C) t(y^2)z
(D) t(y^2)(z^2)
(E) (x^2)(y^2)(z^2)
Rationale: ty > 0 means that ty is positive.
yz < 0 means that yz is negative.
t(y^2)z = (ty)(yz) = positive*negative = negative. Thus option C is always negative.
Answer: (C) t(y^2)z
Question: A certain bacteria colony doubles in size every day for 19 days, a which point it reaches the limit of its habitat and can no longer grow. If two bacteria colonies start growing simultaneously, how many days will it take them to reach the habitat’s limit?
(A) 6.33 
(B) 7.5 
(C) 10 
(D) 15 
(E) 18
Rationale: Case 1: Consider the bacteria num to be B. As it doubles every day, number of bacteria at the end of each day would be
Day 1 – B
Day 2 – 2B
Day 3 – 4B
Day 4 – 8B
.
.
.
Day 19 – 2^19 B
It can be implied that the capacity of the habitat is 2^19 B.
Case 2: There are 2 bacteria which keep doubling every day. So, the count would be
Day 1 – 2B
Day 2 – 4B
Day 3 – 8B
.
.
.
Day 18 – 2^18 B
Day 19 – 2^19 B --> this is the point where the habitat is filled to its capacity.
Hence, 18 is the answer
E
Answer: (E) 18
Question: If two dice are thrown together, the probability of getting an even number on one die and an odd number on the other is?
(A) 1/4
(B) 1/2
(C) 1/8
(D) 1/0
(E) 1/1
Rationale: The number of exhaustive outcomes is 36.
Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then = 18/36 = 1/2
P(E) = 1 - 1/2 = 1/2.
Answer: B
Answer: (B) 1/2
Question: The average of five numbers id 29. If one number is excluded, the average becomes 25. What is the excluded number?
(A) 30
(B) 40
(C) 32.5
(D) 45
(E) 37
Rationale: Explanation:
Sum of 5 numbers = 5 × 29
Sum of 4 numbers after excluding one number = 4 × 25
Excluded number
= 5 × 29 - 4 × 25
= 145 - 100 = 45
Answer: Option D
Answer: (D) 45
Question: John walking at 4 Kmph reaches his office 8 minutes late. If he walks at 6 Kmph, he reaches there 9 minutes earlier. How far is the office from his house?
(A) 3 2/5 Km
(B) 3 1/5 Km
(C) 4 1/5 Km
(D) 5 1/5 Km
(E) None
Rationale: Formula = S1*S2/S2-S1 * T1+T2/60
= 4*6/2 * 17/6
= 24/2 * 17/60
= 17/5 = 3 2/5 Km
A
Answer: (A) 3 2/5 Km
Question: The average weight of 8 person's increases by 3 kg when a new person comes in place of one of them weighing 70 kg. What might be the weight of the new person?
(A) 94 kg
(B) 85 kg
(C) 90 kg
(D) 100 kg
(E) 110 kg
Rationale: Total weight increased = (8 x 3) kg = 24 kg.
Weight of new person = (70 + 24) kg = 94 kg.
Answer : A
Answer: (A) 94 kg
Question: A soccer store typically sells replica jerseys at a discount of 30 percent to 50 percent off list price. During the annual summer sale, everything in the store is an additional 20 percent off the original list price. If a replica jersey's list price is $80, approximately what percent E of the list price is the lowest possible sale price?
(A) 20
(B) 25
(C) 30
(D) 40
(E) 50
Rationale: Let the list price be 2x
for min sale price, the first discount given should be 50%, 2x becomes x here
now, during summer sale additional 20% off is given ie sale price becomes 0.8x
it is given lise price is $80
=> 2x = 80
=> x = 40
and 0.8x = 32
so lowest sale price is 32, which E is 40% of 80
hence, D is the answer
Answer: (D) 40
Question: 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 4 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 is the approximate number of fish in the pond?
(A)  200
(B)  325
(C)  565
(D)  625
(E)  700
Rationale: total fish = x
percentage of second catch = (4/50)*100 = 8%
so, x * 8% = 50
x = 625 ans D.
Answer: (D)  625
Question: in how many ways can a letters of word "BALLOON" is arranged so that two Ls do not come together
(A) 600
(B) 700
(C) 840
(D) 900
(E) 920
Rationale: Total # of arrangements: 7!/[2!*2*]
Total # of arrangements with the LL together:
Note: this means the LL is like one letter.
# = 6!/2!
Therefore, total number with the LL NOT together:
7!/(2!*2!) - 6!/2! = 1260 - 360 = 900
ANSWER:D
Answer: (D) 900
Question: If 625^(-x) + 25^(-2x) + 5^(-4x) = 14, what is the value of x?
(A) -4
(B) -1/4
(C) 0
(D) 1/4
(E) 4
Rationale: We're told that 625^(-X) + 25^(-2X) + 5^(-4X) = 15. We're asked for the value of X.
Since each of the calculated terms MUST be positive (regardless of what the exponent is), we can use thebasesto our advantage.....
With Answer A, we'd have 625^4, which is MUCH BIGGER than 15 (and we'd be adding to that big number). Eliminate A.
With Answer E, we'd have 625^(-4), which would create a TINY fraction (and we'd add some other fractions to it, so the total would be MUCH TOO SMALL). Eliminate E.
With Answer D, we'd have 625^(-1/4), which will also be a fraction (just not as tiny as the one in answer E), but the total would still be TOO SMALL. Eliminate D.
With Answer C, anything to the '0 power' is 1, so we'd have 1+1+1 = 3. This is not 15. Eliminate C.
D
Answer: (D) 1/4
Question: How many different ways can you select a pilot and a co pilot if you have 50 people to choose from?
(A)  199
(B)  200
(C)  398
(D)  2450
(E)  9900
Rationale: You are not just choosing two from 50. Each of the two people chosen fills a particular slot. The first filling the pilot slot and the second filling the copilot slot is different from the first filling the copilot slot and the second filling the pilot slot.
So the answer is in a sense (50C2) x 2 = 2450=D
Answer: (D)  2450
Question: If a, b, c, d, e and f are integers and (ab + cdef) < 0, then what is the maximum number W of integers that can be negative?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Rationale: Minimuum should be 1
Maximum should be 4:
1 out of a or b to make the multiplication negative
3 out of c, d, e or f to make the multiplication negative.
Negative+Negative<0
Answer:C
maximum will be 5..
you dont require both the multiplicatin to be negative for entire equation to be negative...
any one a or b can be negative to make ab negative and it can still be more(away from 0) than the multiplication of 4 other -ve numbers...
actually by writing minimum required as 1 out of 6,you are actually meaning 5 out of 6 also possible as you will see W=5 or 1 will give you same equation..
ans D
Answer: (D) 5
Question: 20 throws of a die produces following results
SCORE -- NUMBER OF OCCURRENCES
---1-------------------4
---2-------------------3
---3-------------------5
---4-------------------2
---5-------------------2
---6-------------------4
What is the probability that one more throw to this series will increase the mean score?
(A) 1/6
(B) 1/3
(C) 1/2
(D) 2/3
(E) 5/6
Rationale: The average score now is \frac{total \ score}{# \ of \ throws}=\frac{1*4+2*3+3*5+4*2+5*2+6*4}{20}=\frac{67}{20}=3.something.
Now, the average score will increase if we get more than the current average, so if we get 4, 5, or 6 on the next throw. The probability of that is 3/6=1/2.
Answer: C.
Hope it's clear.
Answer: (C) 1/2
Question: If there are three sum of money P, Q and R so that P is the simple interest on Q and Q is the simple interest of R, rate % and time are same in each case, then the relation of P, Q and R is given by
(A) P2 = QR
(B) Q2 = PR
(C) R2 = PQ
(D) PQR = 100
(E) None of these
Rationale: P = Q×r×t/100 and Q = R×r×t/100
⇒ P/Q=Q/R=r×t/100
∴ Q2 = PR.
Answer B
Answer: (B) Q2 = PR
Question: A certain lab experiments with white and brown mice only. In one experiment, 2/4 of the mice are white. If there are 16 brown mice in the experiment, how many mice in total are in the experiment?
(A) 32
(B) 36
(C) 34
(D) 39
(E) 42
Rationale: Let total number of mice = M
Number of white mice = 2/4 M
Number of brown mice = 2/4 M = 16
=> M = 32
Answer A
Answer: (A) 32
Question: A certain characteristic in a large population has a distribution that is symmetric about the mean a. If 68% of the distribution lies one standard deviation d of the mean, what percent of the distribution is less than a+d?
(A)  16%
(B)  32%
(C)  48%
(D)  84%
(E)  92%
Rationale: 16%
________________________________________________ a+d
34%
________________________________________________ a
34%
________________________________________________ a-d
16%
since 68% lies one standard deviation from mean a, => 50% of 68% lies on either side as it is symmetric about a.
thus 16% lie below a-d and 16% lie above a+d
now below a+d = 16+34+34 = 84%
hence D
Answer: (D)  84%
Question: For each positive integer,the quantitySn is defined such thatSn+2=(Sn)^4 -S(n+1).in addition ,S2=1.then which of the following must be true.
(A) a)S4 +S3 +1=0
(B) b)(S1)^2 +S4 +2=0
(C) c) ifS1=-1thenS4 is also equal to-1.
(D) d) ifS4 =-7thenS1=-2.
(E) e) ifS1 =-3thenS4=-7
Rationale: Since all the answers include S4, let's start from there.
S4=S2^2-S3
S3=S1^2-S2
Substituting S3 into the first equation we get:
S4=S2^2-(S1^2-S2)
Since S2=1
S4=1-(S1^2-1)
S4=2-S1^2
We need S1^2 in order to calculate S4.
E) provides S1 and from there we can calculate S4:
S4=2-(-3)^2
S4=-7
Correct.
As a side note, I would be very careful with answers in the formif S4=x then S1=y(such as option D ), as we are dealing with S1^2 (not S1), S1 can have two different values and it can not be determined for sure which root is the right answer.
For example:
If S4=-7 then S1=3 :
if we substitute S4 with - 7 we get:
-7=2-S1^2
S1^2=9
now S1 can be +/-3 !!
C
Answer: (C) c) ifS1=-1thenS4 is also equal to-1.
Question: A grocer has a sale of Rs. 3435, Rs. 3920, Rs. 3855, Rs. 4230 and Rs. 3560 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 3500?
(A) 500
(B) 1000
(C) 6000
(D) 7000
(E) 2000
Rationale: Total sale for 5 months = Rs. (3435 + 3920 + 3855 + 4230 + 3560) = Rs. 19000.
Required sale = Rs. [ (3500 x 6) - 19000 ]
= Rs. (21000 - 19000)
= Rs. 2000
Answer: Option E
Answer: (E) 2000
Question: How many such 4 consecutive numbers are there less than 1000 when added gives a sum which is divisible by 10? For example, 16 +17+18+19=70.
(A) 22
(B) 200
(C) 87
(D) 99
(E) 77
Rationale: Explanation:
Let the numbers be n, n+1, n+2, n+3.
Sum of these numbers = 4n + 6.
Let 4n + 6 = 10k
So n = 10k−6410k−64 = 2k−1+2k−242k−1+2k−24 = 2k−1+k−122k−1+k−12
So k takes only odd numbers.
For k = 1, n = 1
For k = 3, n = 6
For k = 5, n = 11
Final value of n = 996.
So total values = =l−ad+1=996−15+1=l−ad+1=996−15+1 = 200
Answer:B
Answer: (B) 200
Question: The radius of a circle is increased by 1%. Find how much % does its area increases?
(A) 2.07%
(B) 2.71%
(C) 2.41%
(D) 2.01%
(E) 2.31%
Rationale: Explanation:
r = 100 r = 101
r2 = 10000 r2 = 10201
10000 ---- 201
100 ---- ? => 2.01%
Answer: D
Answer: (D) 2.01%
Question: The average of 10 numbers is calculated as 15. It is discovered later on that while calculating the average, one number namely 36 was wrongly read as 26. The correct average is?
(A) 12
(B) 26
(C) 16
(D) 97
(E) 12
Rationale: 10 * 15 + 36 – 26
= 160/10
= 16
Answer: C
Answer: (C) 16
Question: David covered a distance of 50 miles on his first trip. On a later trip he traveled 450 miles while going three times as fast. His new time compared with the old time was:
(A) twice as much
(B) three times as much
(C) the same
(D) half as much
(E) a third as much
Rationale: Let t be the time of the first trip.
On the later trip, David can travel 150 miles in time t.
The trip of 450 miles takes 3t.
The answer is B.
Answer: (B) three times as much
Question: An amount of Rs.1000 is divided between 5 persons. Each man gets Rs.250 while each woman gets Rs.125. Find the number of men and women.
(A) 2 men, 3 women
(B) 4 men, 1 woman
(C) 1 man, 4 women
(D) 3 men, 2 women
(E) None of the above
Rationale: Let x be men, y be women
x+y=5.............eq(1)
250x+125y=1000..........eq(2)
solve eq 1 and 2
we get
x=3,y=2
ANSWER:D
Answer: (D) 3 men, 2 women
Question: An art gallery owner is hanging paintings for a new show. Of the five paintings she has to choose from, she can only hang three on the main wall of the gallery. Assuming that she hangs as many as possible on that wall, in how many ways can she arrange the paintings?
(A) 18
(B) 30
(C) 60
(D) 120
(E) 216
Rationale: Arrangment means permutations questions..
5P3 = 5!/3! = 60.
IMO option C is correct answer..
Answer: (C) 60
Question: How many no.'s with 3 different digits, 2, 3, 5, 6, 7 and 9, which are divisible by 5?
(A) 10
(B) 20
(C) 30
(D) 40
(E) 50
Rationale: Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20
B
Answer: (B) 20
Question: Company X provides bottled water to its employees in 5 liter bottles, $3 each. At least how many 5L bottles must Company X buy monthly, so that new contract with $50 fixed monthly rate and $1 for 20L bottle each paid off? (Assume that no other costs apply)
(A)  15
(B)  17
(C)  19
(D)  20
(E)  21
Rationale: let the no. of 5 liter bottles be x, so the no. of 20l bottles will be x/4 (to equate the vol.) since the total cost will be equal , 3x= 50+x/4 so x= 18.18 or 19. Answer is (C).
Answer: (C)  19
Question: sunil start from home at 4 pm and reaches school everyday at 5pm to pick up his children.on saturday the school got over ai 4 pm and children started walking towards home.sunil met them on the way and returned home 15min early.if speed of sunil is 35km/hr then for how many min did children walk? what is speed of children???
(A) 4 km/hr.
(B) 5 km/hr.
(C) 6 km/hr.
(D) 7 km/hr.
(E) 8 km/hr.
Rationale: sunil reach 15 min early. it is time save at both for going to school and coming to home.
so, sunil meet children befor 15/2= 7.5 min.
speed of sunil is 35km/hr i.e. 583.33 m/min.
remaining distance to cover by sunil or else we can say total distance cover by children is 7.5*583.33=4375 m
Sunil meet children after 52.5 min i.e. (1hr-7.5min).
so 4375 is distance cover by children in 52.5min.
so, speed of children is 4375/52.5=83.33m/min=5 km/hr.
ANSWER:B
Answer: (B) 5 km/hr.
Question: A dishonest dealer professes to sell his goods at Cost Price but still gets 20% profit by using a false weight. What weight does he substitute for a kilogram?
(A) 833 1/7 grams
(B) 833 1/3 grams
(C) 833 8/3 grams
(D) 883 1/3 grams
(E) 832 1/3 grams
Rationale: If the cost price is Rs.100, then to get a profit of 20%, the selling price should be Rs.120.
If 120kg are to be sold, and the dealer gives only 100kg, to get a profit of 20%.
How many grams he has to give instead of one kilogram(1000 gm).
120 gm ------ 100 gm
1000 gm ------ ?
(1000 * 100)/120 = 2500/3 = 833 1/3 grams.
Answer:B
Answer: (B) 833 1/3 grams
Question: What number should replace the question mark in this series?
13 44 88 176 847 ?
(A) 1495
(B) 1595
(C) 1695
(D) 1458
(E) 1795
Rationale: First number is 13. Second number is 13+31 = 44. Third number is 44+44=88. Fourth number is 88+88 = 176. Fifth number is 176+671=847. So, next number is 847+748 = 1595
ANSWER:B
Answer: (B) 1595