To find out how many crayons Mrs. White ordered, we first need to calculate the total number of crayons she ordered from both the large and small boxes.
Given:
- Each small box has 90 crayons.
- The number of small boxes ordered = 9.
Total number of crayons from small boxes:
\[ \text{Total Small Crayons} = 9 \text{ boxes} \times 90 \text{ crayons/box} = 810 \text{ crayons} \]
Next, we need to know how many crayons are in each large box to calculate the total for the large boxes. Since the number of crayons in each large box is not specified, we cannot calculate the total crayons from large boxes directly.
If we assume each large box also has 90 crayons (although typically, large boxes contain more), we can calculate it as follows:
Let’s say each large box has \( x \) crayons.
Total for large boxes:
\[ \text{Total Large Crayons} = 6 \text{ boxes} \times x \text{ crayons/box} = 6x \text{ crayons} \]
Now we add the total from both small and large boxes:
\[ \text{Total Crayons} = 810 + 6x \]
Given the choices (1,400; 1,800; 2,100; 3,000), if we hypothesize typical values for large boxes (like 120, 180, or 240), let's solve for when \( 6x + 810 \) equals one of the options.
-
For 1,400 crayons:
\[ 6x + 810 = 1,400 \]
\[ 6x = 590 \]
\[ x \approx 98.33 \] (not an integer) -
For 1,800 crayons:
\[ 6x + 810 = 1,800 \]
\[ 6x = 990 \]
\[ x = 165 \] (possible) -
For 2,100 crayons:
\[ 6x + 810 = 2,100 \]
\[ 6x = 1,290 \]
\[ x = 215 \] (possible) -
For 3,000 crayons:
\[ 6x + 810 = 3,000 \]
\[ 6x = 2,190 \]
\[ x = 365 \] (possible)
So, given common crayon counts, it makes sense that Mrs. White could have ordered based on conventional box sizes.
Assuming 180 crayons in each large box:
Total for 6 large boxes:
\[ 6 \times 180 = 1,080 \]
Total comes out: \[ 1,080 + 810 = 1,890 \] (not in the options)
Upon considering all options and confirming typical crayon quantities in boxes:
The best choice based on calculations and reasonable assumptions is: 1,800 crayons.