This EduRev document offers 10 Multiple Choice Questions (MCQs) from the topic Number System (Level - 3). These questions are of Level - 3 difficulty and will assist you in the preparation of CAT & other MBA exams. You can practice/attempt these CAT Multiple Choice Questions (MCQs) and check the explanations for a better understanding of the topic.
Question for Practice Questions Level 3: Number System - 1
Try yourself:If A * B = (A + B)(A ×^{ }B - A^{2} + B^{2}), then the value of 7 * 5 will be
Explanation
A * B = (A + B)(A × B - A^{2} + B^{2})
7 * 5 = (7 + 5)(35 - 7^{2} + 5^{2})
= (12)(35 - 49 + 25)
= 132
Question for Practice Questions Level 3: Number System - 1
Try yourself:By which of the following numbers must 34300 be multiplied in order to make it a perfect square?
Explanation
34300 = 7 × 7 × 7 × 5 × 5 × 2 × 2 = 7^{3} × 5^{2} × 2^{2}
To make 34300 a perfect square, multiply it by 7.
Question for Practice Questions Level 3: Number System - 1
Try yourself:Find the last digit of (100008)^{12500}.
Explanation
There will be a cycle of last digits, i.e. 8, 4, 2, 6,... and this cycle consists of 4 numbers.
To find the last digit, divide the required power by 4 and find the remainder.
So, when 12,500 is divided by 4, it gives 0 as the remainder.
Therefore, the last digit required would be 6.
Question for Practice Questions Level 3: Number System - 1
Try yourself:How many zeros are there at the end of the product 33 × 175 × 180 × 12 × 44 × 80 × 66?
Explanation
33 = 3 × 11
44 = 2^{2} × 11
175 = 5^{2} × 7
80 = 2^{4} × 5
180 = 5 × 2^{2} × 3^{2}
66 = 2 × 3 × 11
12 = 2^{2} × 3
Multiplying all, we get
2^{11} × 5^{4} × 3^{5} × 7 × 11^{3} = 10^{4} × 2^{7} × 3^{5} × 7 × 11^{3}
Look out for the number of tens, making total number of zeros.
Number of zeros = 4
Question for Practice Questions Level 3: Number System - 1
Try yourself:How many factors of 1260 not end with a zero?
Explanation
1260 = 4 × 315
= 4 × 5 × 63
= 2^{2} × 3^{2} × 5 × 7
The factor has to end with a zero means it should have 2 and 5 necessarily.
So, if we take out 2 and 5, then we will have factors that not ended with 0 - 2 × 3^{2} × 7.
The required answer is the number of factors of 2 × 3^{2} × 7.
⇒ (1 + 1)(2 + 1)(1 + 1)
= 12
Question for Practice Questions Level 3: Number System - 1
Try yourself:If the number 23 x 4534 x 01 is divisible by 11, then how many values can x be substituted with?
Explanation
Sum of the digits at even places = 11 + x
Sum of the digits at odd places = 11 + x
Difference is 0.
i.e. The given number is divisible by 11, irrespective of the value of x.
So, x can be anything from 0 to 9.
Thus, x can be substituted with 10 values.
Question for Practice Questions Level 3: Number System - 1
Try yourself:If the last two digits of a four-digit number are interchanged, the new number obtained is greater than the original number by 54. What is the difference between the last two digits of the number?
Explanation
Suppose that the last two digits of the four-digit number are c and d, where c is at the tens place and d is at the units place.
We can ignore the first two digits.
This gives us,
(10c + d) - (10d + c) = 54
9(c - d) = 54
c - d = 6
Question for Practice Questions Level 3: Number System - 1
Try yourself:How many values can 'n' take, such that 2^{n} is exactly divisible by n^{2}?
Explanation
If n = 1, 2^{1} is divisible by 1^{2}
n = 2, 2^{2} is divisible by 2^{2}
n = 4, 2^{4} is divisible by 4^{2}
n = 8, 2^{8} is divisible by 8^{2}
n = 16, 2^{16} is divisible by 16^{2}
And so on...
So, 'n' can take infinite values, such that 2^{n} is exactly divisible by n^{2}.
Question for Practice Questions Level 3: Number System - 1
Try yourself:A number consists of 3 consecutive digits, such that the digit in the units place being the greatest of the three. The number formed by reversing the digits exceeds the original number by 22 times the sum of the digit. Find the number.
Explanation
Let the hundreds digit be a.
Then, the tens digit = a + 1 and the units digit = a + 2
⇒ The number = 100a + 10(a + 1) + a + 2 = 111a + 12
The number formed by reversing the digits = 100(a + 2) + 10(a + 1) + a
= 111a + 210
Since the number formed by reversing the digits exceeds the original number by 22 times, the sum of the digits
⇒^{ }111a + 210 - 111a - 12 = 22(a + 2 + a + 1 + a)
⇒ 198 = 66a + 66
⇒^{ }a = 2
Hence, the required number = 111a + 12 = 111 × 2 + 12 = 234
Question for Practice Questions Level 3: Number System - 1
Try yourself:Which of the following is a multiple of 88?
Explanation
Only 1,43,616 is a multiple of both 11 and 8 and so of 88.
For a number to be divisible by '8' the last three digits should be multiple of '8'.
Option (1) 13,92,578 the last three digits i.e. 578 is not divisible by 8.
Option (2) 1,38,204 the last three digits i.e. 204 is not divisible by 8.
Option (3) 14,36,240 the last three digits i.e. 240 is divisible by 8.
Option (4) 1,43,616 the last three digits i.e. 616 is divisible by 8.
Option (5) 1,43,661 the last three digits i.e. 661 is not divisible by 8.
Now check option (3) and option (4) for multiple of '11'.
For 14,36,280 the sum of digits at odd places and sum of digits at even places is 0 + 2 + 3 + 1 = 6 and 8 + 6 + 4 = 18.
Now 18 - 6 = 12 not multiple of 11. Hence, number 14,36,280 is not multiple of 88.
For 1,43,616 the sum of digits at odd places and sum of digits at even places is 6 + 6 + 4 = 16 and 1 +3 + 1 = 5.
Now 16 -5 = 11.
Hence, 1,43,616 is multiple of 88.