Question 1 :

What is the value of square root of 81?

(A)  81              (B)  3               (C)  9

Solution :

√81  =  √(3 ⋅ 3 ⋅ 3 ⋅ 3)

=  3

=  9

Question 2 :

Find the area of a right angled triangle, if the radius of the circumcircle is 5 cm and altitude to the hypotenuse is 4 cm.

(A)   20 cm2              (B)   10 cm2           (C)   30 cm2

Solution :

Let ABC be the right angled triangle right angled at B. Let O be the center of the circumcircle.

Then, O is the midpoint of the hypotenuse AC.

OA  =  OB  =  OC (radius of circle 5 cm)

Hypotenuse AC  =  Diameter of the circle

=  2  5  =  10 cm

Let BM be the perpendicular from B on AC.

MB  =  4 cm

Area of right angled triangle ABC

=  (1/2)  Base  Height

=  (1/2)  10  4

=  20 cm2

So, the area of the right triangle is 20 cm2.

Question 3 :

The volume of cubical box is 46.656 cubic meters. Find the length of the side of the cubical box.

(A)   3.6 m              (B)   5.2 m           (C)   7.8 m

Solution :

Volume of cubical box  =  46.656 cubic meters

a3  =  46.656

a  =  3.6

So, side length of cubical box  =  3.6 meter

Question 4 :

The sum of the digits of a two digit number is 15 and if 9 is added to the number the digits are interchanged. Find the required number.

(A)  78          (B)   87         (C) 96

Solution :

Let "xy" be the two digit number

Sum of the two digit number  =  15

x + y  =  15  -----(1)

If 19 is added to the number the digits are interchanged.

xy + 9  =  yx

10x + 1y + 9  =  10y + 1x

10x - x + 1y - 10y  =  -9

9x - 9y  =  -9  -----(2)

 (1)⋅9  ==> 9x+9y  =  135                9x-9y  =  -9           ___________               18x  =  126                x  =  7 Applying the value of x in the first equation, we get7 + y  =  15Subtract 7 on both sides,y  =  15 - 7  y =  8

So, the required two digit number is 78.

Question 5 :

In a circle with center O and radius 17 cm, PQ is a chord at a distance of 8 cm from the center. Find the length of the chord.

(A)  30 cm     (B)  15 cm     (C)  24 cm

Solution :

In a right triangle OCQ,

OQ2  =  OC2 + CQ2

172  =  82 + CQ2

289 - 64  = CQ2

CQ  =  225  ==>  CQ  =  15

PQ  =  2 CQ

=  2(15)  =  30 cm

So, the length of chord is 30 cm.

Question 6 :

The radius and height of a cylinder are in the ratio 5:7 and its volume is 550 cm³ find its radius.

(A)  5 cm     (B)  6 cm     (C)  7 cm

Solution :

r : h  =  5 : 7

r/h  =  5/7

r  =  5h/7

Volume of cylinder  =  550 cm³

∏r2 h  =  550

(22/7) ⋅ (5h/7)2 h  =  550

h3  =  550 ⋅ (7/22) (49/25)

h3  =  343

h  =  7

r  =  5

So, the radius is 5 cm.

Question 7 :

The mean weight of 4 boys is 56 kg and that of 6 girls in 46 kg. find the combined mean weight of 10 students.

(A)  40 kg     (B)  36 kg     (C)  50 kg

Solution :

Mean weight of 4 boys  =  56 kg

Weight of 4 boys  =  56 (4)  =  224

Mean weight of 6 girls  =  46 kg

Weight of 6 girls  =  46 (6)  =  276

Total weight  =  224  +  276

=  500

Average weight  =  500/10  =  50 kg

So, the average weight is 50 kg.

Question 8 :

The arithmetic mean of 8, 10, x and 12 is 9. Find the value of x.

(A)  6                (B)   8               (C)   9

Solution :

Arithmetic mean  =  9

(8 + 10 + x + 12)/4  =  9

30 + x  =  9(4)

30 + x  =  36

Subtract 30 on both sides

x  =  36 - 30

x  =  6

So, the value of x is 6.

Question 9 :

Find the number of sides of regular polygon whose exterior angle measure of 45 degree.

Solution :

Sum of the exterior angles of regular polygon  =  360°

But each exterior angle  =  45°

number of sides of regular polygon  =  360° / 45°

=  8

So, the number of sides of the regular polygon is 8.

Question 10 :

Simplify the expressions and evaluate them as directed.

x (x-3) + 2 for x = 1

Solution :

x (x - 3) + 2

x  =  1

=  1 ( 1 - 3) + 2

=  1 (-2) + 2

=  -2 + 2

=  0

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