mensuration Model Questions & Answers, Practice Test for ssc steno grade c d 2024

Answer: (b)

Radius of sphere (r) = 9 cm = 0.09 m

Diameter of wire (d) = 0.4 cm

⇒ R = 0.2 cm = 0.002 m

According to question,

Volume of sphere = Volume of wire

⇒ $4/3 πr^3 = πR^2h$

∴ h = $4/3 × r^3/R^2$

= $4/3 × {0.09 × 0.09 × 0.09}/{0.002 × 0.002}$ = 81 × 3 = 243 m

Answer: (b)

Here, we produced AB line to M.

Since, AM is parallel to CD.

∠DCM = ∠BMC = x (alternate angle)

Also, ABM is a straight line.

∠EBM = π – y

Now in ΔBEM.

∠B + ∠M +∠E = π

⇒ π – y + x + ∠E = π

⇒ ∠E = y – x

Answer: (a)

Let radius and height of cylinder be 2x and 3x

Respectively

$π r^2$ h = 1617

$π (2x)^2$ 3x = 1617

π $12x^3$ = 1617

$x^3 = {1617 × 7}/{22 × 12} = (7/2)^3$

x = $7/2$ ; r = 7 h = ${21}/2$

Total surface area = 2πr(r + h)

= ${2 × 22}/7 × 7[7 + {21}/2]$

⇒ $44 [{35}/2] ⇒ 22 × 35 ⇒ 770 cm^2$

Answer: (c)

We are given that a rectangular cello tape of length 4 cm and breadth 0.5 cm is used for joining each pair of edges. Therefore, area of the cello tape used for each face of the cube is $a^2$ . Thus, total area of the cello tape used is $6a^2$. Now, we have $6a^2$ = 6(4 × 0.4)(4 × 0.5) = 6.2.2 = 24 sq. cm.

Answer: (c)

Required speed of flow of water

= ${225 × 162 × 20}/{5 × 100} = {60}/{100} × {45}/{100} × h$

∴ h = 5400