Advanced maths Quiz For SSC CGL -1

#everydayquiz #10Questions

1.    What will be the length of the diagonal of that square plot whose area is equal to the area of a rectangular port of length 45 meters and breadth 40 meters?

(a) 42.5 meters                 

(b) 60 meters

(c) Data inadequate           

(d) 45 meters

2.    The length of a rectangle is 20% more than its breadth. What will be the ratio of the area of a rectangle to that of a square whose side is equal to the breadth of the rectangle?

(a) 2:1                                

(b) 6:5

(c) 3:4                                

(d) Data inadequate

3.    If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.

(a) 40 cm                          

(b) 50 cm

(c) 60 cm                          

(d) 30 cm

4.    The base of triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.

(a) 200 m                          

(b) 275 m

(c) 300 m                          

(d) 600 m

5.    The difference between two parallel sides of a trapezium is 4 cm. The perpendicular distance between them is 19 cm. If the area of the trapezium is 475 cm², find the lengths of the parallel sides.

(a) 27 cm & 23 cm           

(b) 22 cm & 25cm

(c) 20 cm & 27 cm           

(d) 23 cm & 25 cm

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 cm²) 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 cm²                        

(b) 33 1/3 cm²

(c) 40 cm²                        

(d) 50 cm²

Answer with solution:

1.   (b) 

Area= (45* 40) m² = 1800 m^2,

(Diagonal)²/2= 1800,

Diagonal = 60m

2.  (b)

  Let breadth be x meters. Then length= 120% of x = 120/100 x= 6x/5 m

Therefore, the required ratio= {(6x/5)*x*x}/x*x = 6:5

3.  (b) 

Let x and y be the length and breadth of the rectangle respectively.

Then, x ­- 4 =y + 3 or x – y = 7 ………..(i)

Area of rectangle = x*y,

Area of square = (x-4) * (y+3)

So, (x-4) * (y+3)= x*y,

or 3x – 4y = 12……………………………….(ii)

Solving eq (i) & (ii) x= 16 and y = 9

Therefore perimeter of rectangle = 2(x+ y)= 2(16+9)= 50 cm

4.  (c) 

 Area of the field = Total cost/ Rate = 333.18/ 24.68 hectares = 13.5 hectares

= (13.5 * 1000) m² = 135000 m²

Let altitude = x meters and base = 3x meters.

Then, 1/2 * 3x * x = 135000

x^{2 =} 90000

x= 300 m

5.   (a) 

Let the two parallel sides of the trapezium be a cm and b cm,

Then, a-b = 4……………………….(i)

And, 1/2 *(a+b)*19 = 475

(a+b) = (475*2)/19 =50…………(ii)

Solving eq(i)& (ii) a= 27 cm and b = 23 cm

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

b² -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 = [6² + 8² + (10)² + (19)² + (20)²]

= (36 + 64 + 100 + 361 + 400) cm²= 961 cm²

Side of the new square = √961 cm= 31 cm

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

8.    (d)

A₁  / A₂ =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) cm²= 6 cm²

Area of required triangle= (1/4 * 6) cm² = 3/2 cm²

10.  (a)

Let the sides be a cm and 2a cm

Then, a² + (2a)² = (10)²

5 a²= 100

a² = 20 cm²

Area = (1/2 *a *2a) = a²= 20 cm²

#SSC #IBPS #SBI #RBI #NABARD #NICL #NIACL #CAT #NMAT #everydayquiz