Numbers in general form | Playing with Numbers | Assessment | English | Grade 8

Generalised form of a three-digit number pqr is ________________.

• 100p + 10r + q

• 100p + 10q + r

• p + 10q + 100r

• p + q + r

Answer explanation

The general form of number pqr is 100p + 10q + r.

Tags

CCSS.2.NBT.A.3

The usual form of the number 1000 × 9 + 100 × 5 + 10 × 7 + 1 × 3 is ________.

• 9753

• 9573

• 9375

• 3579

Answer explanation

The correct usual form for 1000a + 100b + 10c + d is abcd. Hence, the usual form for 1000 × 9 + 100 × 5 + 10 × 7 + 1 × 3 is 9573.

Tags

CCSS.4.NBT.A.2

When written in general form, the number 12xyz is equal to _______________.

• 12 + x + y + z

• 10000 × z + 1000 × y + 100 × x + 10 × 2 + 1 × 1

• 10000 × 1 + 1000 × 2 + 100 × x + 10 × y + 1 × z

• 12 + 100 × x + 10 × y + 1 × z

Answer explanation

The correct general form of the number 12xyz is 10000 × 1 + 1000 × 2 + 100 × x + 10 × y + 1 × z

Tags

CCSS.4.NBT.A.2

Choose the correct usual form of (a + 100b + 1000c) - (100q + 1000 p + 10r)

• abc - qpr

• abc - pqr0

• cb0a - pqr

• cb0a - pqr0

Answer explanation

Let's find the usual form of the first term (a + 100b + 1000c) = 1000c + 100b + 10 × 0 + a = cb0a Let's find the usual form of second term (100q + 1000 p + 10r) = 1000p + 100q + 10r + 1 × 0 = pqr0 Therefore, (a + 100b + 1000c) - ( 100q + 1000 p + 10r) = cb0a - pqr0 So Option 4 is the correct answer.

Tags

CCSS.4.NBT.B.4

If the general form of the number 20912 is 10000 × 2 + a + 1000 × b + 100 × 9 + 10 × c, then identify the values of a, b and c.

• a = 2, b = 1, c = 0

• a = 0, b = 9, c = 2

• a = 0, b = 1, c = 0

• a = 2, b = 0, c = 1

Answer explanation

Given number is 20912 The general form is : 10000 × 2 + a + 1000 × b + 100 × 9 + 10 × c Let's rearrange the given general form : 10000 × 2 + 1000 × b + 100 × 9 + 10 × c + a Now let's write general form of the number 20912 : 10000 × 2 + 1000 × 0 + 100 × 9 + 10 × 1 + 2 By comparing both the general forms, we get, a = 2, b = 0, c = 1

Tags

CCSS.4.NBT.A.2