**Hello philic's,**

**Many aspirants told us to start daily quiz on the advanced section of maths for ssc exam, keeping their Dream on mind, we started this section, we'll update it on daily basis. As SSC CGL exam is coming near, many of candidates are studying very hard for this exam and want to crack this exam very seriously. Many of candidates understand this thing that QUANTITATIVE APTITUDE Section plays a pivotal role in deciding the cut-off for candidates giving this exam especially the ADVANCED MATHS portion.**

__Quant Quiz For SSC (advance Maths) part-2__

**1**.A cube of edge 5 cm is cut into cubes of each edge 1 cm. The ratio of the total surface area of one of the small cubes to that of the large cube is equal to

A.1:5

B.1:25

C.1:125

D.1 : 625

**2**.The cost of painting the whole surface area of a cube at the rate of 13 paise per Sq.cm is Rs. 343.98. Then the volume of the cube is

A.8500cm^3

B.9000cm^3

C.9250cm^3

D.9261cm^3

**3.**A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm when a man gets on it. The mass of man is

A.12kg

B.60kg

C.72kg

D.96kg

**4.**How many bricks, each measuring 25 cm x 11.25 cm x 6 cm, will be needed to build a wall of 8 m x 6 m x 22.5 cm?

A.5600

B.6000

C.6400

D.7200

**5.**A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:

A.720

B.900

C.1200

D.1800

**6.**A farmer wishes to start a 100sq m triangular vegetable garden. Since he has only 30m barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. The dimension of the garden is.

(a) 15 m * 6.67
m

(b) 20 m * 5 m

(c) 30 m *
3.33m

(d) 40m * 2.5 m

**7.**The perimeter of 5 squares is 24 cm, 32 cm, 40 cm, 76 cm and 80 cm respectively. The perimeter of another square equal in area to the sum of the areas of these squares is:

(a) 31
cm

(b) 62 cm

(c) 124
cm

(d) 961 cm

**8.**If the ratio of areas of two squares is 225:256, then the ratio of their perimeters is:

(a)
256:225

(b) 225:256

(c)
16:15

(d) 15:16

**9.**The sides of a triangle are 3 cm, 4 cm, and 5 cm. The area (in cm

^{2}) of the triangle formed by joining the mid-points of the sides of this triangle is:

(a)
3/4

(b) 3/2

(c) 3

(d) 6

**10.**One of sides of a right-angled triangle is twice the other, and the hypotenuse is 10 cm. The area of the triangle is :

(a) 20 cm

^{2 }

(b) 33 1/3 cm

^{2}
(c) 40 cm

^{2 }
(d) 50 cm

^{2}**Download KD Campus SSC CGL Tier-1 Mock Tests Free.**

__Answer with complete solution:__
1.(6x1x1 / 6x5x5)

‹=› 1/ 25

‹=› 1 : 25.

2.Surface area = (34398 / 13)

‹=›2646cm^3

‹=›6a^2= 2646

‹=›a^2= 441

‹=›a = 21.

So,volume =(21x21x21)cm^3= 9261cm^3

3.Volume of water displaced = (3 x 2 x 0.01)m^3

= 0.06m3.

Mass of man =Volume of water displaced ×Density of water

= (0.06×1000) kg

= 60 kg.

4.Number of bricks = Volume of the wall/Volume of 1 brick

=(800x600x22.5 /25x11.25x6)

= 6400

5. 2(15 + 12) ×h = 2(15 x 12)

‹=›h= 180/27 m

= 20/3m

Volume = (15×12×20/3)m^3

‹=›1200 m^3.

‹=› 1/ 25

‹=› 1 : 25.

2.Surface area = (34398 / 13)

‹=›2646cm^3

‹=›6a^2= 2646

‹=›a^2= 441

‹=›a = 21.

So,volume =(21x21x21)cm^3= 9261cm^3

3.Volume of water displaced = (3 x 2 x 0.01)m^3

= 0.06m3.

Mass of man =Volume of water displaced ×Density of water

= (0.06×1000) kg

= 60 kg.

4.Number of bricks = Volume of the wall/Volume of 1 brick

=(800x600x22.5 /25x11.25x6)

= 6400

5. 2(15 + 12) ×h = 2(15 x 12)

‹=›h= 180/27 m

= 20/3m

Volume = (15×12×20/3)m^3

‹=›1200 m^3.

^{}**6.**(b)

According to quest, 2b + L = 30, (as fencing is only for 3 sides)

So, L =30 – 2b

Now area of the garden = 100 sq m,

L * b =100, OR b*(30-2b) = 100

b

^{2}-15b+ 50 =0,
now b= 5 & b=10

if b = 5 then L = 20 and if b= 10 then L= 10

since the garden is rectangular so the dimension would be 20m * 5 m

**7.**(c)

The sides of the 5 squares are (24/4), (32/4), (40/4), (76/4), (80/4) i.e., 6 cm, 8 cm , 10 cm , 19 cm, 20 cm

Area of the new square = [6

^{2}+ 8^{2}+ (10)^{2}+ (19)^{2}+ (20)^{2}]
= (36 + 64 + 100 + 361 + 400) cm

^{2}= 961 cm^{2}
Side of the new square = √961 cm= 31 cm

Perimeter of the new square = (4*31) cm = 124 cm

**8.**(d)

A

_{1 }/ A_{2 }=225/256 = (15)^{2}/(16)^{2}
=15/16

Now 4 A1 / 4A2= 15/16

**9.**(b)

A= 3 cm, b= 4 cm, and c= 5 cm

It is a right-angled triangle with base = 3 cm and height= 4 cm

So its area= (1/2 * 3 * 4) cm

^{2}= 6 cm^{2}
Area of required triangle= (1/4 * 6) cm

^{2 }= 3/2 cm^{2}**10.**(a)

Let the sides be a cm and 2a cm

Then, a

^{2}+ (2a)^{2 }= (10)^{2}
5 a

^{2}= 100
a

^{2 }= 20 cm^{2}
Area = (1/2 *a *2a) = a

^{2}= 20 cm^{2}
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