Today we look at the detailed solutions for Free test 001 for CAT, XAT and other MBA Exams.
1. 5 Students take a math test, and score different marks. The sum of the scores taken 2 at a time are: 38, 51, 55, 60, 64, 68, 72, 77, 85, 94. What is the difference between student with highest and lowest score?
Let a,b,c,d,e be the 5 scores, with a < b < c < d < e.
So, a + b = 38, d + e = 94 -- (1)
Also, adding all scores 4(a+b+c+d+e) = 664
Or, a + b + c + d + e = 166 -- (2)
From (1) and (2),
c = 166 - 38 - 94 = 34
The second largest number, 51, has to be the sum of a,c and second largest number, 85, has to be the sum of c,e
So, a = 51 - 34 = 17 and e = 85 - 34 = 51
Required difference = 51 - 17 = 34.
2. What is the remainder when 7100 is divided by 121?
73 = 243 = 1(mod 121)
100 = 3x33 + 1
So, 7100= 133x7 (mod 121) = 7(mod 121)
3. Which among the following is the smallest?
21/2, 31/3, 41/4, 61/6 and 121/12
We will start by converting all numbers to the 12th power:
21/2 = 26/12 = 641/12
31/3 = 34/12 = 811/12
41/4 = 43/12 = 641/12
61/6 = 62/12 = 361/12
121/12 = 121/12
So, we have the smallest number as 121/12
4. The sum of four consecutive two-digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers?
The last digits of 4 consecutive odd numbers can be
(1,3,5,7) = last digit will be 6
(3,5,7,9) = last digit will be 4
(5,7,9,1) = last digit will be 2
(7,9,1,3) = last digit will be 0
(9,1,3,5) = last digit will be 8
Of this, (7,9,1,3) is the only possibility.
Let the numbers be 10a+7, 10a+9, 10a+11, 10a+13, where a is a single digit number.
Sum = 40a + 40 = 10 (4a+4)
Dividing by 10, we have 4(a+1)
This means that a+1 is a perfect square, and a > 0
So, a = 3 or 8
We have: 37,39,41,43 and 87,89,91,93.
Of these only 41 is present in the options.
5. A number of children are standing in a circle. They are evenly spaced and the 14th child is diametrically opposite the 37th child. How many children are in the circle?
Since the 14th is diametrically opposite the 37th, there are 37 - 14 = 23 children in the semicircle
So total children = 23x2 = 46
1. 5 Students take a math test, and score different marks. The sum of the scores taken 2 at a time are: 38, 51, 55, 60, 64, 68, 72, 77, 85, 94. What is the difference between student with highest and lowest score?
Let a,b,c,d,e be the 5 scores, with a < b < c < d < e.
So, a + b = 38, d + e = 94 -- (1)
Also, adding all scores 4(a+b+c+d+e) = 664
Or, a + b + c + d + e = 166 -- (2)
From (1) and (2),
c = 166 - 38 - 94 = 34
The second largest number, 51, has to be the sum of a,c and second largest number, 85, has to be the sum of c,e
So, a = 51 - 34 = 17 and e = 85 - 34 = 51
Required difference = 51 - 17 = 34.
2. What is the remainder when 7100 is divided by 121?
73 = 243 = 1(mod 121)
100 = 3x33 + 1
So, 7100= 133x7 (mod 121) = 7(mod 121)
3. Which among the following is the smallest?
21/2, 31/3, 41/4, 61/6 and 121/12
We will start by converting all numbers to the 12th power:
21/2 = 26/12 = 641/12
31/3 = 34/12 = 811/12
41/4 = 43/12 = 641/12
61/6 = 62/12 = 361/12
121/12 = 121/12
So, we have the smallest number as 121/12
4. The sum of four consecutive two-digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers?
The last digits of 4 consecutive odd numbers can be
(1,3,5,7) = last digit will be 6
(3,5,7,9) = last digit will be 4
(5,7,9,1) = last digit will be 2
(7,9,1,3) = last digit will be 0
(9,1,3,5) = last digit will be 8
Of this, (7,9,1,3) is the only possibility.
Let the numbers be 10a+7, 10a+9, 10a+11, 10a+13, where a is a single digit number.
Sum = 40a + 40 = 10 (4a+4)
Dividing by 10, we have 4(a+1)
This means that a+1 is a perfect square, and a > 0
So, a = 3 or 8
We have: 37,39,41,43 and 87,89,91,93.
Of these only 41 is present in the options.
5. A number of children are standing in a circle. They are evenly spaced and the 14th child is diametrically opposite the 37th child. How many children are in the circle?
Since the 14th is diametrically opposite the 37th, there are 37 - 14 = 23 children in the semicircle
So total children = 23x2 = 46
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