Geometry Solved Qs
Also, some important Qs I am posting for my own reference which I feel can be a big time drainer in the test.
Q. If 60! is written out as an integer, with how many consecutive 0’s will that integer end?
6
12
14
42
56
Ans: If an integer ends in 0, then that integer is divisible by 10; when that integer is divided by 10, the final 0 disappears. Therefore, the number of 0’s at the end of an integer is exactly the number of times that integer is divisible by 10.
Since 10 = 2 × 5, this number (the number of times the integer is divisible by 10) is the smaller of the powers to which 2 and 5 are raised in the integer’s prime factorization. For instance, an integer whose prime factorization contains 28 × 57 and an integer whose prime factorization contains 27 × 58 will both end with seven 0’s.
In a factorial, such as 60!, there are many more 2’s than 5’s in the prime factorization: the factorial product has more numbers contributing 2’s than 5’s, and more of those numbers contribute repeated factors. Therefore, the question can be rephrased: How many 5’s are in the prime factorization of (60!) ?
60! is the product of all the integers from 1 to 60, inclusive. Each of the twelve multiples of 5 in this range – 5, 10, 15, …, 60 – contributes a 5 to the prime factorization. Additionally, a second 5 is contributed by each of the two multiples of 25 (25 and 50). Therefore, the prime factorization of 60! contains fourteen 5’s, and so 60! will end with fourteen 0’s.
The correct answer is C.
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NOTE on Quadratic Eqn :(y – 4)(y + 1) = 0
This quadratic has two possible solutions for y: 4 and -1. HOWEVER, you must check these solutions in the original equation, since squaring both sides of an equation can introduce a spurious, invalid solution.
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Q. If 4^(4x) = 1600, what is the value of (4 ^(x–1)2?
It is tempting to express both sides of the equation 4^(4x) = 1600 as powers of 4 and to try and solve for x. That is a catch GMAT offers to waste our time. As soon as something like this appears in the test, we should try to utilize given data directly into what the Q is asking and find out the value rather than solving for x.
Q.In a certain game, a large bag is filled with blue, green, purple and red chips worth 1, 5, x and 11 points each, respectively. The purple chips are worth more than the green chips, but less than the red chips. A certain number of chips are then selected from the bag. If the product of the point values of the selected chips is 88,000, how many purple chips were selected?
1
2
3
4
5
Ans:88,000 is the product of an unknown number of 1's, 5's, 11's and x's. To figure out how many x’s are multiplied to achieve this product, we have to figure out what the value of x is. Remember that a number's prime box shows all of the prime factors that when multiplied together produce that number: 88,000's prime box contains one 11, three 5’s, and six 2’s, since 88,000 = 11 × 53 × 26.
The 11 in the prime box must come from a red chip, since we are told that 5 < 6 =" 2" 10 =" 2" 8 =" 23">GMAT prep -- Prime Factor
1. between 2 and 10
2. between 10 and 20
3. between 20 and 30
4. between 30 and 40
5. greater than 40
H(100) is product of all even integers from 2 to 100
H(100)= 2*4*6*8....96*98*100
This exp can be re-written as
H(100)=(2*1)*(2*2)*(2*3)*(2*4).....*(2*48)*(2*49)* (2*50)
Taking factor of 2 common out of every term in the bracket
so, H(100)= [2*2*2..50 times] *(1*2*3..49*50)
H(100)=2^50(1*2*3..*48*49*50)
now what are the prime factors of h(100)?
it is: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47
So, that means h(100) is divisible by all the prime factors listed above.
now question is asking the smallest prime factor of h(100)+1,. i.e. a prime number that divides h(100)+1 completely.
Now, as we know that h(100) is divisible by all prime factors less than or equal to 47, so if we divide h(100)+1 by any prime factor less than or equal to 47, we will always get remainder as 1.
This implies that there is no prime number less than or equal to 47 that will divide h(100)+1 completely. Hence, the smallest prime number that will divide h(100)+1 HAS to be greater than 47. The moment I know this, I mark E and move on.
Q:methods to calculate total average speed
Say, a car travels at S1 mph on a trip and at S2 mph on return trip. What is its average speed for the entire trip?
Solution:
*** Don't fall in the trap of just averaging the 2 speeds. Overall average speed is not (S1+S2)/2. ***
Total average speed is simply = Total distance/Total time
Lets say,
D = distance travelled by the car in EACH direction
t1 = time spent on onward trip
t2 = time spent on return trip
Thus, the total distance travelled by the car = D+D= 2D
And, by the formula, Speed = Distance/Time
S1 = D/t1 => t1 = D/S1
S2 = D/t2 => t2 = D/S2
Total average speed = Total Distance/Total time = 2D/(t1+t2) = 2D/(D/S1+D/S2) = 2S1*S2/(S1+S2)
Example:
A car travels at 60 mph on a trip and at 100 mph on return trip. What was its average speed for the entire trip?
Solution:
*** Total average speed is not (60+100)/2 = 80 ***
Total average speed = 2*60*100/(100+60) = 2*60*100/160 = 2*60*5/8 = 60*5/4 = 15*5 = 75
Data sufficiency
Q:Is x < z? (1) x = (1/2)y (2) y = 0.5z
Answer:From (1) AND (2) : If x = (1/2)y, and y = 0.5z, then x = (1/2)(0.5z) = (1/2)(1/2)z = (1/4)z.
If z is positive, then x is less than z. For example: If z = 4, then x = 1.
If z is negative, then x is greater than z. For example: If z = -4, x = -1.
Thus, INSUFFICIENT
The correct answer is E.
Standard Deviation, generally asked Qs tests below concept in DS,
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the standard deviation of a set is a measure of how far data points in a distribution fall from the mean. Therefore, standard deviation can equal zero only when all the elements in the set are equal to the mean, i.e. when all the elements in the set are the same.
Palindrome Problem
A palindrome is a number that reads the same forward and backward, such as 121. How many odd, 4-digit numbers are palindromes?
Answer:
Notice that the outer two numbers must match and the inner two numbers must match, creating numbers such as 1221 or 6556. Since the last digit must be odd, our only choices are 1, 3, 5, 7, or 9 for the first/last digit. There are no restrictions on the inner digits, so we have 10 choices, 0 – 9.
Since we have 5 choices for the outer two digits and 10 choices for the inner two digits, using the counting principle we have 5 × 10 = 50 choices for our 4-digit number. Notice that we do not set the problem up as 5 10 10 5 and multiply, giving 2500. There are only two choices to be made – number of possibilities for inner digits and number of possibilities for outer digits.
The correct answer is 50.
Terminating Decimal Question and Rule
Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals.
| If a, b, c, d and e are non-negative integers and p = 2a3b and q = 2c3d5e, is | p  q | a terminating decimal? |
(1) a > c
Answer:
For fraction p/q to be a terminating decimal, the numerator must be an integer and the denominator must be an integer that can be expressed in the form of 2x5y where x and y are integers. (Any integer divided by a power of 2 or 5 will result in a terminating decimal.)
Remainder Problem Trick
If m and n are positive integers such that m/n = 13.24, which of the following could NOT be the remainder when m is divided by n?
 | 12 | |
| | 16 | |
| | 30 | |
| | 48 | |
| | 120 |
This question is about remainders. Whenever we have a remainder problem, it might be useful to think in terms of the following framework. Assume that x and y are integers, I is the quotient, and R is the remainder:
| x | = I + | R |
It should be noted that R and I will always be integers, and R will always be greater than or equal to zero, and always less than y. For example:
| 13 | = 2 + | 3 |
When we divide 13 by 5, we get a quotient of 2 and a remainder of 3. Note that in this case, the remainder is indeed larger than (or equal to) zero, and less than 5.
Also note that if R = 0, then x is divisible by y, and vice versa.
| Let’s apply this formula to the problem at hand. We know that | m | = 13.24. This tells us that: |
| m | = 13.24 = I + | R |
| Since I is the quotient, I = 13. That tells us that .24 = | R | , or .24n = R |
Because 0.18 is a terminating decimal and n and R are integers, it might be easier to convert 0.18 to a fraction:
0.24n = R
| 24 | n = R |
| 6 | n = R |
Now, let’s move the denominator to the other side of the equation:
6 × n = 25 × R
What does that tell us about n and R? Well, we know that 6, n, 25, and R are all integers. Thus both sides of the equation will be integers (the same integer). For that to be true, both sides of the equation must have IDENTICAL prime factorizations.
We know that the left side of the equation has a 2 and a 3 in its prime factorization (6 = 2 × 3). Therefore, R must have at least a 2 and a 3 in its prime factorization. So R is divisible by 6. Furthermore, we know that the right side of the equation has two 5’s in its prime factorization (25 = 5 × 5). Therefore, n must have at least a 5 and a 5 in its prime factorization. So n is divisible by 25.
The question asks about R. We know that R must be divisible by 6. Only answer choice B, 16, is not divisible by 6.
DS Problem
What is the distance between x and y on the number line?
(1) |x| – |y| = 5
(2) |x| + |y| = 11
Answer : One tip on this problem is that whenever you are asked about the distance between two points on a number line, it is mathematically equivalent to the absolute value of the difference between those two points. Thus in the problem, the question can be rephrased as:
What is |x – y|?
Answer is Both statement together not sufficient to prove..i.e. E
Perfect Square
The sum of the factors of any perfect square is odd.
If N is a perfect square, then N has an odd number of factors.
Absolute Value Property
If |X+Y| = |X|+|Y|
Either, X > 0, Y > 0 or X <0,Y<0
Imply, XY > 0
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