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Question: 14 One-third, one-fourth, one-fifth and one-seventh of the human population of Island X, which has fewer than 5000 human inhabitants, are all whole numbers & their sum is exactly the population of island Y. What is the population of Island Y?
Option: (a) 4200 (b) 4279 (c) 4581 (d) 4800 (e) None of these
Explore: Since 1/3rd, 1/4th, 1/5th & 1/7th of the population are all whole numbers, hence, the total population must be a multiple of the LCM of 3, 4, 5 & 7.
LCM of 3, 4, 5, 7 = 3x4x5x7 = 12x35 = 420
Now, 5000/420 11(20/21). So, the largest number less than 5000, that is also a multiple of
420 is 420x11 = 4620
:. Population of Y = 4620 x (1/3 + 1/4 + 1/5 + 1/7) = 4620((105+140+84+62)/420) = 11x389 = 4279
Answer: (b)
Question: 15 Find the value of k if (X + 1) is a factor of X3+ kx + 3x3- 2.
Option: (a) 3.5 (b) 4 (c) 4.5 (d) 5 (e) None
Explore: Theoretically, we will have to determine what value of k makes (X3+ kx + 3x3- 2)/(x+1) an integer.
But doing so is very tough & is not possible within the time limit of the exam. So, we will try to plug in the values from the answer and see if it works.
First let, k = 4
:. X3+ kx + 3x3- 2
= X3+ 3x + 3x2 + 1 - 3 +x
This is not divisible by (x + 1)
Let, k = 5
:. X3+ kx + 3x3- 2
= X3+ 3x + 3x2 + 1 - 3 + 2x
= (x+1)3+ 2x - 3
This too is not divisible by (x+1)
Even by trying k = 3.5 and k = 4.5 you will see that you cannot factorize it. So, (e) is the answer.
Question: 16 Suppose z = a x c x d x e where a > b > c > d > e. A decrease by 1 in which of the factors would result in the greatest decrease in the value of z?
Option: (a) a (b) b (c) c (d) d (e) e
Explore: To solve this problem, you think of multiplication as repeated addition. So, a reduction of 1 in any of the factors will reduce 1 addition. To make the maximum decrease, therefore, we have to reduce the addition of the largest term by 1.
The largest possible combination of 4 variables is given by a x b x c x d [:.a > b > c > d > e]
So, we need to reduce e by 1 to get the greatest decrease in the value of z.
Now let us check this by using numbers.
e.g 6 > 5 > 4 > 3 > 2
6x5x4x3x2 = 720
6x5x4x3x1 = 360
6x5x4x2x2 = 480
6x5x3x3x2 = 540
6x4x4x3x2 = 576
5x5x4x3x2 =600
So, you see, reducing 2 to 1 gives the maximum reduction.
Answer: (e)
Question: 17 Which of the following is a factor of 12?
Option: (a) 24 (b) 40 (c) 18 (d) 4
Explore: Answer: (d) Factors of 12:2, 3, 4, 6, 12
Question: 18 Which of the following numbers is divisible by 2, 3, 5, 7?
Option: (a) 42 (b) 105 (c) 420 (d) 30 (e)None
Explore: Answer: (c) LCM of 2, 3, 5, 7 = 2x3x5x7 = 10x21 = 210
Since 420 = 2x210, So C is true.
Question: 19 Which of the following integers has the most divisors?
Option: (a) 88 (b) 91 (c) 95 (d) 99 (e) 101
Explore: 88 = 2x2x2x11
Its factors are 1, 2, 4, 8, 11, 22, 44, 88 = 8
91 = 7x13 :. Its factors are 1, 7, 13, 91 = 4 factors
95 = 5x19 :. Its factors are 1, 5, 95 = 3 Factors
99 = 3x3x11 :. Its factors are 1, 3, 9, 11, 33, 99 = 6 Factors
101 = 1x101 :. Its factors are 1, 101 = 2 factors
Answer: (a)
Please also check our first and second post:
Option: (a) 4200 (b) 4279 (c) 4581 (d) 4800 (e) None of these
LCM of 3, 4, 5, 7 = 3x4x5x7 = 12x35 = 420
Now, 5000/420 11(20/21). So, the largest number less than 5000, that is also a multiple of
420 is 420x11 = 4620
:. Population of Y = 4620 x (1/3 + 1/4 + 1/5 + 1/7) = 4620((105+140+84+62)/420) = 11x389 = 4279
Answer: (b)
Question: 15 Find the value of k if (X + 1) is a factor of X3+ kx + 3x3- 2.
Option: (a) 3.5 (b) 4 (c) 4.5 (d) 5 (e) None
Explore: Theoretically, we will have to determine what value of k makes (X3+ kx + 3x3- 2)/(x+1) an integer.
But doing so is very tough & is not possible within the time limit of the exam. So, we will try to plug in the values from the answer and see if it works.
First let, k = 4
:. X3+ kx + 3x3- 2
= X3+ 3x + 3x2 + 1 - 3 +x
This is not divisible by (x + 1)
Let, k = 5
:. X3+ kx + 3x3- 2
= X3+ 3x + 3x2 + 1 - 3 + 2x
= (x+1)3+ 2x - 3
This too is not divisible by (x+1)
Even by trying k = 3.5 and k = 4.5 you will see that you cannot factorize it. So, (e) is the answer.
Question: 16 Suppose z = a x c x d x e where a > b > c > d > e. A decrease by 1 in which of the factors would result in the greatest decrease in the value of z?
Option: (a) a (b) b (c) c (d) d (e) e
Explore: To solve this problem, you think of multiplication as repeated addition. So, a reduction of 1 in any of the factors will reduce 1 addition. To make the maximum decrease, therefore, we have to reduce the addition of the largest term by 1.
The largest possible combination of 4 variables is given by a x b x c x d [:.a > b > c > d > e]
So, we need to reduce e by 1 to get the greatest decrease in the value of z.
Now let us check this by using numbers.
e.g 6 > 5 > 4 > 3 > 2
6x5x4x3x2 = 720
6x5x4x3x1 = 360
6x5x4x2x2 = 480
6x5x3x3x2 = 540
6x4x4x3x2 = 576
5x5x4x3x2 =600
So, you see, reducing 2 to 1 gives the maximum reduction.
Answer: (e)
Question: 17 Which of the following is a factor of 12?
Option: (a) 24 (b) 40 (c) 18 (d) 4
Explore: Answer: (d) Factors of 12:2, 3, 4, 6, 12
Question: 18 Which of the following numbers is divisible by 2, 3, 5, 7?
Option: (a) 42 (b) 105 (c) 420 (d) 30 (e)None
Explore: Answer: (c) LCM of 2, 3, 5, 7 = 2x3x5x7 = 10x21 = 210
Since 420 = 2x210, So C is true.
Question: 19 Which of the following integers has the most divisors?
Option: (a) 88 (b) 91 (c) 95 (d) 99 (e) 101
Explore: 88 = 2x2x2x11
Its factors are 1, 2, 4, 8, 11, 22, 44, 88 = 8
91 = 7x13 :. Its factors are 1, 7, 13, 91 = 4 factors
95 = 5x19 :. Its factors are 1, 5, 95 = 3 Factors
99 = 3x3x11 :. Its factors are 1, 3, 9, 11, 33, 99 = 6 Factors
101 = 1x101 :. Its factors are 1, 101 = 2 factors
Answer: (a)
Please also check our first and second post:
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