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This mock test of Olympiad Test: Number System - 2 for Class 7 helps you for every Class 7 entrance exam.
This contains 20 Multiple Choice Questions for Class 7 Olympiad Test: Number System - 2 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

Which of the following is the unit digit in {(6374)^{1793} × (625)^{317 }× (341^{491})}?

Solution:

Unit digit in (6374)^{1793} = Unit digit in (4)^{1793}

= Unit digit in [(42)^{896} × 4]

= Unit digit in (6 × 4) = 4

Unit digit in (625)^{317} = Unit digit in (5)^{317} = 5

Unit digit in (341)^{491} = Unit digit in (1)^{491 }= 1

Required digit = Unit digit in (4 × 5 × 1) = 0.

QUESTION: 2

The last unit digit in the expansion of given expression 71^{n}–1, where n is any positive integer is:

Solution:

The unit digit in the 71^{n} for any value of n is 1.

Thus the unit digit is 71^{n} –1 is 0.

QUESTION: 3

Find the value of:

Solution:

Given Expression =

QUESTION: 4

What is the value of given number expression? 107 x 107 + 93 × 93 = ?

Solution:

107 × 107 + 93 × 93 = (107)^{2} + (93)^{2}

= (100 + 7)^{2} + (100 – 7)^{2}

= 2 × [(100)^{2} + 7^{2}] [Ref: (a + b)^{2} + (a – b)^{2}

= 2(a^{2} + b^{2})] = 20098

QUESTION: 5

Which of the following is the remainder when we divide (67^{67} + 67) by 68?

Solution:

(x^{n} + 1) will be divisible by (x + 1) only when n is odd.

(67^{67} + 1) will be divisible by (67 + 1)

(67^{67} + 1) + 66, when divided by 68 will give 66 as remainder.

QUESTION: 6

Find the sum of first 5 prime numbers.

Solution:

Required sum = (2 + 3 + 5 + 7 + 11) = 28.

QUESTION: 7

Replace question mark with the suitable answer:

(12345679 × 2 × 6^{2}) = ?

Solution:

(12345679 × 2 × 6^{2}) = 12345679 × 72

= 12345679 × (70 +2)

= 12345679 × 70 + 12345679 × 2

= 864197530 + 24691358

= 888888888

QUESTION: 8

Find the value of n if (64)^{2} – (36)^{2 }= 20 × n.

Solution:

20 × n = (64 + 36)(64 – 36) = 100 × 28

QUESTION: 9

Which of the following is the correct value of given number expression: (22 + 4^{2} + 6^{2} + ... + 20^{2}) = ?

Solution:

(2^{2} + 4^{2} + 6^{2} + ... + 20^{2})

= (1 × 2)^{2} + (2 × 2)^{2} + (2 × 3)^{2} + ... + (2 × 10)^{2}

= (2^{2} × 1^{2}) + (2^{2} × 2^{2}) + (2^{2} × 3^{2}) + ...

+ (2^{2} × 10^{2})

= 2^{2} × [1^{2} + 2^{2} + 3^{2} + ... + 10^{2}]

[Formula: (1^{2} + 2^{2} + 3^{2} + ... + n^{2})

= (4 × 5 × 77)

= 2 × 77 × 10

QUESTION: 10

What will be the sum of first forty five counting numbers?

Solution:

Let S_{n} = (1 + 2 + 3 + ... + 45)

This is an A.P. in which a = 1, d = 1, n = 45 and l = 45

Required sum = 207 × 5.

QUESTION: 11

What will be the sum of even numbers between 1 and 21 is?

Solution:

Let S_{n} = (2 + 4 + 6 + ... + 20).

This is an A.P.

in which a = 2, d = 2 and l = 20

Let the number of terms be n. Then,

a + (n – 1)d = 20

2 + (n – 1) × 2 = 20

n = 10.

QUESTION: 12

What is the value obtained if we multiply 2056 and 987?

Solution:

2056 × 987 = 2056 × (1000 – 13)

= 2056 × 1000 – 2056 × 13

= 2056000 – 26728

= 2029272

QUESTION: 13

If two-third of three-fourth of a number is 30, then three – fifth of that number is:

Solution:

Let the number be x.

QUESTION: 14

What is the value of third integer if 3 times the first of 3 odd consecutive integers is three more than twice the third?

Solution:

Let the three integers be x, x + 2 and x + 4.

Then, 3x = 2(x + 4) + 3

x = 11

Third integer = x + 4 = 15

QUESTION: 15

Find the value of given number expression: 397 × 397 + 104 × 104 + 2 × 397 × 104 = ?

Solution:

Given number expression = (397)^{2} + (104)^{2} + 2 × 397 × 104

= (397 + 104)^{2}

= (501)2 = (500 + 1)^{2}

= (500)2 + (1)^{2} + (2 × 500 × 1)

= 250000 + 1 + 1000

= 251001

QUESTION: 16

Which of the following value is true in place of question mark? (35423 + 7164 + 41720) – (317 × 89) = ?

Solution:

(35423 + 7164 + 41720) – (317 × 89)

= (35423 + 7164 + 41720) – (28213)

= 84307 – 28213

= 56094

QUESTION: 17

(a^{n} – b^{n}) is completely divisible by (a – b), when

Solution:

For every natural number n, (a^{n} – b^{n}) is completely divisible by (a – b).

QUESTION: 18

Find the divisor by which the given number expression (3^{25} + 3^{26} + 3^{27} + 3^{28}) is completely divided.

Solution:

(3^{25} + 3^{26} + 3^{27} + 3^{28}) = 3^{25} × (1 + 3 + 32 + 33)

= 3^{25} × 40

= 3^{24} × 3 × 4 × 10

= (3^{24} × 4 × 30), which is divisible by 30.

QUESTION: 19

When we divide a number say x by 6 it leaves a remainder 3. When we divide the square of the same number by 6, what will be the remainder?

Solution:

Let x = 6q + 3.

Then, x^{2} = (6q + 3)^{2}

= 36q^{2} + 36q + 9

= 6(6q^{2} + 6q + 1) + 3

Thus, when x^{2} is divided by 6, then remainder = 3.

QUESTION: 20

Which of the following is the remainder when we divide 17^{200} by 18?

Solution:

When n is even; (x^{n}- a^{n}) is completely divisible by (x + a)

(17^{200 }– 1^{200}) is completely divisible by (17 + 1), i.e., 18.

(17^{200} – 1) is completely divisible by 18.

On dividing 17^{200} by 18, we get 1 as remainder.

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