mensuration area volumes Model Questions & Answers, Practice Test for ssc cgl tier 1 2023

Answer: (c)

Radius of the inner track = 100 m

and time = 1 min 30 sec ≡90 sec.

Also, Radius of the outer track = 102 m

and time = 1 min 32 sec ≡ 92 sec.

Now, speed of A who runs on the inner track

= ${2 π (100)}/{90} = {20 π}/{9}$ = 6.98

And speed of B who runs on the outer track

= ${2 π (102)}/{90} = {51 π}/{23}$ = 6.96

Since, speed of A > speed of B

∴ A runs faster than B.

Answer: (b)

Area between square and semi-circles

= Area of square – 2x Area of semi-circle

$(10)^2 - 2x {100}/7 × {(5)^2}/2 = 100 - 78.5 = 21.5 cm^2$

Answer: (d)

Volume of solid cylinder = π$(3)^2{4} = 36π cm^3$

and volume of conical cavity

= $1/3 π{(3)^2}(4) = 12 π cm^3$

∴ Volume of remaining solid

= 36π – 12π = 24π $cm^3$

Answer: (d)

In ΔABC, by phythagoras

BC = $√{p^2 + q^2}$

Area of the semicircle with diameter BC

= $π/2 ({√{p^2 + q^2}}/2)^2$

= $π/8 (p^2 + q^2)$

Area of the semicircle with diameter AB

$π/2 (p/2)^2 = {π p^2}/8$

Area of the semicircle with diameter AC

= $π/2 (q/2)^2 = {π q^2}/8$

Area of the triangle ABC = $1/2$ pq.

Area of shaded region =

${πp^2}/8 + {π q^2}/8 + 1/2 pq - π/8 (p^2 + q^2)$

= ${π (P^2 + q^2)}/8 + 1/2 pq - π/8 (p^2 + q^2)$

= $1/2$ pq.

Answer: (c)

According to question,

27 × Volume of smaller drops = Volume of bigger drop

∴ $27 × 4/3 π r^3 = 4/3 π R^3$

⇒ 27 × $(0.2)^3 = R^3$

⇒ $(3 × 0.2)^3 = R^3$

R = 0.6 cm