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Test Level 3: Number System - 1 - CAT MCQ


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10 Questions MCQ Test - Test Level 3: Number System - 1

Test Level 3: Number System - 1 for CAT 2024 is part of CAT preparation. The Test Level 3: Number System - 1 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 3: Number System - 1 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 3: Number System - 1 below.
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Test Level 3: Number System - 1 - Question 1

If A * B = (A + B)(A × B - A2 + B2), then the value of 7 * 5 will be

Detailed Solution for Test Level 3: Number System - 1 - Question 1

A * B = (A + B)(A × B - A2 + B2)
7 * 5 = (7 + 5)(35 - 72 + 52)
= (12)(35 - 49 + 25)
= 132

Test Level 3: Number System - 1 - Question 2

By which of the following numbers must 34300 be multiplied in order to make it a perfect square?

Detailed Solution for Test Level 3: Number System - 1 - Question 2

34300 = 7 × 7 × 7 × 5 × 5 × 2 × 2 = 73 × 52 × 22
To make 34300 a perfect square, multiply it by 7.

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Test Level 3: Number System - 1 - Question 3

Find the last digit of (100008)12500.

Detailed Solution for Test Level 3: Number System - 1 - Question 3

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.

Test Level 3: Number System - 1 - Question 4

How many zeros are there at the end of the product 33 × 175 × 180 × 12 × 44 × 80 × 66?

Detailed Solution for Test Level 3: Number System - 1 - Question 4

33 = 3 × 11
44 = 22 × 11
175 = 52 × 7
80 = 24 × 5
180 = 5 × 22 × 32
66 = 2 × 3 × 11
12 = 22 × 3
Multiplying all, we get
211 × 54 × 35 × 7 × 113 = 104 × 27 × 35 × 7 × 113
Look out for the number of tens, making total number of zeros.
Number of zeros = 4

Test Level 3: Number System - 1 - Question 5

How many factors of 1260 not end with a zero?

Detailed Solution for Test Level 3: Number System - 1 - Question 5

1260 = 4 × 315
= 4 × 5 × 63
= 22 × 32 × 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 × 32 × 7.
The required answer is the number of factors of 2 × 32 × 7.
⇒ (1 + 1)(2 + 1)(1 + 1)
= 12

Test Level 3: Number System - 1 - Question 6

If the number 23 x 4534 x 01 is divisible by 11, then how many values can x be substituted with?

Detailed Solution for Test Level 3: Number System - 1 - Question 6

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.

Test Level 3: Number System - 1 - Question 7

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?

Detailed Solution for Test Level 3: Number System - 1 - Question 7

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

Test Level 3: Number System - 1 - Question 8

How many values can 'n' take, such that 2n is exactly divisible by n2?

Detailed Solution for Test Level 3: Number System - 1 - Question 8

If n = 1, 21 is divisible by 12
n = 2, 22 is divisible by 22
n = 4, 24 is divisible by 42
n = 8, 28 is divisible by 82
n = 16, 216 is divisible by 162
And so on...
So, 'n' can take infinite values, such that 2n is exactly divisible by n2.

Test Level 3: Number System - 1 - Question 9

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.

Detailed Solution for Test Level 3: Number System - 1 - Question 9

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

Test Level 3: Number System - 1 - Question 10

Which of the following is a multiple of 88?

Detailed Solution for Test Level 3: Number System - 1 - Question 10

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.

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