**Sixth Grade Math Practice Online :**

In this section, we will see some practice questions for 6th grade students.

**Question 1 :**

The sum of numbers divisible by 7 between 50 and 100 ?

(A) 539 (B) 600 (C) 425

**Solution :**

Numbers which are divisible by 7 between 50 and 100 are

56, 63, 70, 77, 84, 91, 98

Now we have to find the sum of these numbers

For that we have to add these numbers

56 + 63 + 70 + 77 + 84 + 91 + 98 = 539

So the answer is 539.

**Question 2 :**

A bag contains red, green and blue balls. The total number of balls is 80. If 60% of the total balls are blue, what is the probability of taking a blue ball?

(A) 1/6 (B) 5/8 (C) 3/5

**Solution :**

Total number of balls n (S) = 80

60% of the total balls are blue

Number of blue balls = 60% of 80

= (60/100) ⋅ 80

= (4800/100)

= 48

Let A be the event of getting blue balls

n(A) = 48

Probability of getting blue balls = n(A)/n(S)

= 48/80

= 3/5

So the answer is 3/5.

**Question 3 :**

√2 is

(A) Rational (B) Irrational (C) Real

**Solution :**

Let √2 be rational

Since √2 is rational we shall write it as fraction

√2 = a/b

Taking squares on both sides

2 = a^{2}/b^{2}

2b^{2} = a^{2}

2 divides a^{2}

2 divides a

let a = 2 c

2b^{2} = (2c)^{2}

2b^{2} = 4c^{2}

c^{2} = 2b^{2}/4

c^{2} = b^{2}/2

2 divides b

from this we come to know that a and b are having common factor 2 other than 1. So we shall decide that a/b is not rational number.

It is irrational.So, √2 is irrational

**Question 4 :**

The distance between A to B is 5 cm, B to C is 6 cm, C to D is 9 cm, D to E is 12 cm, E to F is 6.5 cm and F to G is 4.8 cm. Then what is the distance between B to F?

(A) 30.4 cm (B) 33.5 cm (C) 12.5 cm

From this we need to find the distance between the points B and F

BF = BC + CD + DE + EF

BC = 6 cm, CD = 9 cm, DE = 12 cm, EF = 6.5 cm

BF = 6 + 9 + 12 + 6.5

BF = 33.5 cm

So the answer is 33.5 cm

**Question 5 :**

The greatest common factor of x² y² z², x² y z² and x y z² is

(A) xy^{2}z^{2 } (B) xy^{2}z (C) xyz^{2}

**Solution :**

x^{2} y^{2} z^{2}, x^{2} y z^{2}, x y z^{2}

To find GCF, we have to select the common terms only

GCF = x y z^{2}

So the correct answer is x y z^{2}.

**Question 6 :**

The sum of a number and its square is 90. Find the required numbers.

(A) 9 and 81 (B) 45 and 45 (C) 20 and 70

**Solution :**

Let x be the required number and x^{2} be its square

x + x^{2} = 90

x + x^{2} - 90 = 0

(x + 10) (x - 9) = 0

x + 10 = 0 x - 9 = 0

x = -10 x = 9

So, the answer is 9 and 81.

**Question 7 :**

In how many ways can five people sit in a round table?

(A) 60 (B) 150 (C) 120

**Solution :**

The number of arrangements in a round table is (n - 1)!

= (5 - 1)!

= 4!

= 24

Hence the number of ways is 24.

**Question 8 :**

If two dice are rolled, what is the probability of getting same number on both the dies?

(A) 1/2 (B) 1/3 (C) 1/6

**Solution :**

Sample space

S = { {1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4)(5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4)(6, 5) (6, 6)}

n(S) = 36

Let A be the event of getting the same number on both the dies

A = { (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6) }

n(A) = 6

p(A) = n(A)/n(S)

= 6/36

= 1/6

So, the required probability is 1/6.

**Question 9 :**

Compare 4/5 and 5/6

(A)(5/6) < (4/5) (B) (5/6) > (4/5) (C) (5/6) = (4/5)

**Solution :**

Since the denominators are not same, first we have to make the denominators same.

L.C.M of 5 and 6 = 30

= (4/5) ⋅ (6/6)

= (24/30)

= (5/6) ⋅ (5/5)

= (25/30)

Therefore 4/5 is less than 5/6

So the answer is (5/6) > (4/5).

**Question 10 :**

Evaluate :

25! / 24!

(A) 25 (B) 24 (C) 1/24

**Solution :**

25! / 24! = 25(24!)/24!

= 25

Hence the value is 25.

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