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Q:

Find the area of the rectangle or square with the given dimensions s=10 feet

A) 500	B) 100
C) 200	D) 300

Answer & Explanation Answer: B) 100

Explanation:

$A = S^{2}$

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Q:

Find the area of the rectangle or square with the given dimensions L = 5 inches, W= 3 inches?

A) 13	B) 14
C) 15	D) 16

Answer & Explanation Answer: C) 15

Explanation:

We know that,

Area of rectangle A = L x W = 5 x 3 = 15 sq.inches.

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Q:

Find the perimeter of the rectangle or square with the given dimensions s = 3 cm

A) 12	B) 13
C) 14	D) 15

Answer & Explanation Answer: A) 12

Explanation:

P=4s

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Q:

Find the perimeter of the rectangle or square with the given dimensions l = 5 feet, w = 2 feet?

A) 11ft	B) 13ft
C) 14ft	D) 15ft

Answer & Explanation Answer: C) 14ft

Explanation:

We know that,

Perimeter of rectangle = 2(L+W) = 2(5 + 2) = 2 (7) = 14 fts.

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Q:

A sector of 120, cut out from a circle, has an area of $\frac{66}{7} c m^{2}$ . Find the radius of the circle ?

A) 1	B) 2
C) 3	D) 4

Answer & Explanation Answer: C) 3

Explanation:

Let the radius of the circle be r cm. Then,

$\frac{120}{360} {πr}^{2} = \frac{66}{7}$

$r^{2} = \frac{66}{7} * \frac{7}{22} * 3$

r=3.

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Q:

Two concentric circles form a ring. The inner and outer circumferences of ring are $\frac{518}{7}$ m and $\frac{352}{7}$ m respectively. Find the width of the ring.

A) 4	B) 5
C) 6	D) 7

Answer & Explanation Answer: A) 4

Explanation:

Let the inner and outer radii be r and R metres.

Then $2 πR = \frac{352}{7} = > R = \frac{352}{7} * \frac{7}{22} * \frac{1}{2} = 8 m$

$2 πR = \frac{528}{7} = > R = \frac{528}{7} * \frac{7}{22} * \frac{1}{2} = 12 m$

=> Width of the ring = (R - r) = (12 - 8) m = 4 m.

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Q:

In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?

A) 15	B) 10
C) 25	D) 20

Answer & Explanation Answer: C) 25

Explanation:

The toys are different; The boxes are identical
If none of the boxes is to remain empty, then we can pack the toys in one of the following ways
a. 2,2,1
b. 3,1,1

Case a. Number of ways of achieving the first option 2−2–1

Two toys out of the 5 can be selected in $5 C_{2}$ 5 ways. Another 2 out of the remaining 3 can be selected in $3 C_{2}$ ways and the last toy can be selected in $1 C_{1}$ way.

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2.

Therefore, total number of ways of achieving the 2−2–1 option is:

$\frac{5 C_{2} * 3 C_{2}}{2} = \frac{10 * 3}{2} = 15$ ways.

Case b. Number of ways of achieving the second option 3−1–1

Three toys out of the 5 can be selected in $5 C_{3}$ ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.
Therefore, total number of ways of getting the 3−1–1 option is $5 C_{3}$ =10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes
=number of ways of achieving Case a + number of ways of achieving Case b
=15 + 10 = 25 ways

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Q:

Find the length of the altitude of an equilateral triangle of side $3 \sqrt{3}$ cm.

A) 4.5	B) 3.5
C) 2.5	D) 6.5

Answer & Explanation Answer: A) 4.5

Explanation:

Let the length of the altitude be h.
Then, h = $a \frac{\sqrt{3}}{2}$ = $3 \sqrt{3} \times \frac{\sqrt{3}}{2}$ = 4.5

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