Question 1 :

In a school the total enrollment of class 8th is 115. If the number of boys exceeds the number of girls by 33, find the number of boys in a class 8th.

(A) 74            (B) 89            (C) 50

Solution :

Let x be the number of girls

Number of boys  =  x + 33

Total number of students in the class  =  115

x + x + 33  =  115

2x + 33  =  115

Subtract 33 on both sides

2x  =  115 - 33

2x  =  82

Divide by 2 on both sides

x  =  41

x + 33  =  74

So, the number of boys in the class is 74.

Question 2 :

One of the angles of a triangles is equal to the sum of the remaining tow angles. If the ratio of these angles is 4:5, find the angles of the given triangle.

(A) 60, 60, 60        (B) 60, 80, 40          (C) 40, 50, 90

Solution :

Let the three angles be 9x, 4x and 5x

Sum of three angles  =  180°

4x + 5x + 4x + 5x  =  180°

8x + 10x  =  180°

18x  =  180°  ==>  x  =  10°

Hence the angles are 90, 50, 40.

Question 3 :

Two equal sides of a triangle are each 4 m less than three times the third side.Find the dimensions of a triangle,if its perimeter is 55 m.

(A) 15, 15, 8              (B) 23, 23, 9         (C) 13, 13, 8

Solution :

Let "x" be the third side of the triangle

Length of two equal sides  =  3x - 4

Perimeter of triangle  =  55 m

3x - 4 + 3x - 4 + x  =  55

7x - 8  =  55

7x  =  63

Divide both sides by 7

x  =  9

3x - 4  =  3(9) - 4  ==> 27 - 4  ==> 23

So,the required angles are 23, 23, and 9

Question 4 :

If a + 1/a = 2, find the value of a3 + (1/a)3

(A) 1            (B) 2             (C) 3

Solution :

To find the value of a3 + (1/a)3, we may use the formula given below.

(a + b)3  =  a3 + b3 + 3ab(a + b)

a3 + a1/3  =  (a + (1/a))+ 3 a (1/a)(a + (1/a))

=  23 + 3 (2)

=  8 + 6

=  14

Question 5 :

A gardener bought twice the number of Lilly plants that he had in his garden but had to throw 3 bad plants. When the new Lilly plants were planted, he had in all 48 plants in the garden.Find the number of Lilly plants he had originally.

(A) 17                 (B) 15               (C) 13

Solution :

Let x be the number of lilly plants in his garden

Number of plants he buys  =  2x

2x + x - 3  = 48

3x - 3  =  48

3x  =  48 + 3

3x  =  51

Divide both sides by 3

x  =  51/3  =  17

So, the number of Lilly plants in the garden is 17.

Question 6 :

Simplify a3 - b3 ÷ (a-b)

(A) a+ ab + b2          (B) a- ab + b2         (C) (a+b) (a+b)

Solution :

a3 - b=  (a - b) (a+ ab + b2)

a3 - b3 ÷ (a - b)  =  (a - b) (a+ ab + b2) / (a - b)

=  a+ ab + b2

Question 7 :

If a + b + c = 9 and ab + bc + ca = 23, find the value of a2 + b2 + c2

(A) 15            (B) 12               (C) 35

Solution :

(a + b + c)2  =  a2 + b2 + c+ 2ab + 2bc + 2ca

(a + b + c)2  =  a2 + b2 + c2  + 2(ab + bc + ca)

a2 + b2 + c2 = (a + b + c)2 - 2(ab + bc + ca)

= (9)2 - 2 (23)

=  81 - 46  =  35

Question 8 :

The difference between the compound interest and the simple interest on a certain principal for 2 years at 4% per annum is \$150. Find the principal

(A) 1875           (B) 3750           (C) 1263

Solution :

Simple interest  =  PNR/100

150  =  P(2)(4) / 100

150  =  8 P / 100

Multiply both sides by 100

150 (100)  =  8 P

Divide both sides by 8

P  =  150(100) / 8

=  1875

So, the principal amount is 1875.

Question 9 :

The watch is marked at \$1150. During off season a discount given and it is sold for \$1100. Find the discount percent allowed.

(A) 2.14%          (B) 4.34%            (C) 3.04%

Solution :

Discount percentage

=  (Discount amount / Marked price) ⋅ 100

Discount amount  =  1150  -  1100  =  50

=  (50/1150) ⋅ 100

=  0.0434 (100)

=  4.34%

Question 10 :

What must be added to each of the numerator and the denominator of the fraction 7/11 to make it equal to 3/4

(A) 2             (B) 8             (C) 5

Solution :

Let x be the required number to be added to both numerator and denominator.

(7 + x) / (11 + x)  =  3/4

4 (7 + x)  =  3 (11 + x)

28 + 4x  =  33 + 3x

4x - 3x  =  33 - 28

x  =  5

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