## Gurdeep is organizing his sunflower seeds, petunia seeds,

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### Gurdeep is organizing his sunflower seeds, petunia seeds,

by AAPL » Thu Nov 07, 2019 11:58 am
Manhattan Prep

Gurdeep is organizing his sunflower seeds, petunia seeds, and daffodil seeds by volume. If the combined volume of all three seeds is 66 cm$$^3$$, what is the volume of his daffodil seeds?

1) The volume of the petunia seeds is 10 percent of the combined volume of sunflower and daffodil seeds.
2) The volume of the daffodil seeds is 20 percent of the combined volume of sunflower and petunia seeds.

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by deloitte247 » Sat Nov 09, 2019 3:21 pm
Let the volume of petunia seed = p
Let the volume of sunflower seed = s
Let the volume of daffodil seed = d
$$p+s+d=66cm^3$$
What is the volume of Gurdeep daffodil seeds (d)?
Statement 1: The volume of petunia seeds is 10 percent of the combined volume of sunflower and daffodil seeds.
$$p=0.1s+0.1d$$
$$p+s+d=66cm^3$$
$$0.1s+0.1d+s+d=66cm^3$$
$$1.1s+1.1d=66cm^3$$
$$d=\frac{66-1.1s}{1.1}$$
The value of s is unknown; hence, statement 1 is NOT SUFFICIENT.

Statement 2: The volume of the daffodil seed is 20 percent of the combined volume of sunflower and petunia seeds.
$$d=0.2\cdot\left(s+p\right)$$
$$d=0.2s+0.2p$$
$$From\ p+s+d=66cm^3$$
$$p=66-s-d$$
$$d=0.2s+0.2\left(66-s-d\right)$$
$$d=0.2s+13.2-0.2s-0.2d$$
$$d+0.2d=13.2$$
$$1.2d=13.2$$
$$d=\frac{13.2}{1.2}=11cm^3$$

Therefore, statement 2 alone is SUFFICIENT.

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### Re: Gurdeep is organizing his sunflower seeds, petunia seeds,

by GMATGuruNY » Wed Jan 20, 2021 1:26 pm
AAPL wrote:
Thu Nov 07, 2019 11:58 am
Manhattan Prep

Gurdeep is organizing his sunflower seeds, petunia seeds, and daffodil seeds by volume. If the combined volume of all three seeds is 66 cm$$^3$$, what is the volume of his daffodil seeds?

1) The volume of the petunia seeds is 10 percent of the combined volume of sunflower and daffodil seeds.
2) The volume of the daffodil seeds is 20 percent of the combined volume of sunflower and petunia seeds.
Statement 1:
p : (s+d) = 10:100 = 1:10
Of every 11 cm^3, p = 1 cm^3 and s+d = 10 cm^3, implying that s+d constitutes 10/11 of the total volume.
No way to determine the value of d.
INSUFFICIENT.

Statement 2:
d : (s+p) = 20:100 = 1:5
Of every 6 cm^3, d = 1 cm^3 and s+p = 5 cm^3, implying that d constitutes 1/6 of the total volume:
d = 1/6 * 66 = 11 cm^3
SUFFICIENT.