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This mock test of Olympiad Test: Number System - 3 for Class 7 helps you for every Class 7 entrance exam.
This contains 10 Multiple Choice Questions for Class 7 Olympiad Test: Number System - 3 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

What will be the value of (1^{2} + 2^{2} + 3^{2} + ... + 10^{2}) = ?

Solution:

We know that (1^{2} + 2^{2} + 3^{2} + ... + n^{2})

Putting n=10, required sum

= 385

QUESTION: 2

What will be the sum of (51+ 52 + 53 + ... + 100) = ?

Solution:

This is an A.P. in which a = 51, l = 100 and n = 50.

QUESTION: 3

What will be the last digit in (7^{95}– 3^{58})?

Solution:

Unit digit in 7^{95} = Unit digit in [(7^{4})^{23} × 7^{3}]

= Unit digit in [(Unit digit in(2401))^{23} × (343)]

= Unit digit in (1^{23} × 343) = Unit digit in (343) = 3

Unit digit in 358 = Unit digit in [(3^{4})^{14} × 3^{2}]

= Unit digit in [Unit digit in (81)^{14} × 3^{2}]

= Unit digit in [(1)^{14} × 3^{2}]

= Unit digit in (1 × 9)

= Unit digit in (9) = 9

Unit digit in (7^{95} – 3^{58})

= Unit digit in (343 – 9)

= Unit digit in (334) = 4.

QUESTION: 4

Find the value of 1904 × 1904 = ?

Solution:

904 × 1904 = (1904)^{2}

= (1900 + 4)^{2}

= (1900)^{2} + (4)^{2} + (2 × 1900 × 4)

= 3610000 + 16 + 15200

= 3625216

QUESTION: 5

The least six digit number completely divisible by 111 is:

Solution:

The least six digit number is 100000.

When 100000 ÷ 111,

Quotient 990 and Remainder = 100

Therefore required number

=100000 + ( 111 – 100)

= 100000 + 11

= 100011

QUESTION: 6

Find the number of terms in the given G.P. series 3, 6, 12, 24... 384.

Solution:

Here a = 3 and r =6/3 =2.

Let the number of terms be n.

Then, t_{n} = 384

ar^{n-1} = 384

3 × 2^{n – 1} = 384

2^{n-1} = 128 = 2^{7}

n – 1 = 7

n = 8

Number of terms = 8.

QUESTION: 7

What will be the largest 5 digit number completely divisible by 91?

Solution:

Largest 5-digit number = 99999

When 99999÷91, Quotient = 1098 and

Remainder = 81

Required number = (99999 – 81)

= 99918.

QUESTION: 8

When we multiply a certain number by 7, we obtain product whose each digit is 3. What will be that number?

Solution:

By hit and trial, we find that

47619 × 7 = 333333.

QUESTION: 9

If 25% of 2/5 of a certain number is 125 then the required number is:

Solution:

Let the required number be n,

Then 25% × 2/5 × n = 125

By solving the above equation,

∴ n = 1250

QUESTION: 10

Which of the following cannot be the square of counting number?

Solution:

The square of a natural number never ends in 7.

42437 is not the square of a natural number.

### Number System (Part - 3)

Video | 08:17 min

### Number System (Part - 3)

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### Ch.1 NUMBER SYSTEM. Lecture 3

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