MATH PRACTICE QUESTIONS FOR GRADE 7 WITH ANSWERS

Question 1 :

Let a > 0 and (x+1)(x+2) + 1 = ax² + bx + c. The values of a, b and c are

(A) (1, 5, 5)        (B) (1, 3, 3)            (C) (0, 1, 5)

Solution :

(x+1)(x+2) + 1  =  ax² + bx + c

x² + 3x + 2 + 1  =  ax² + bx + c

x² + 3x + 3  =  ax² + bx + c

a  =  1, b  =  3 and c  =  3.  

Question 2 :

If ten players participate in a tennis tournament in which each player plays every other player exactly once. After each match, the two players shake hands. Then, both players shake hands with the umpire. After all of the matches, how many handshakes have been exchanged?

(A) 115           (B) 155         (C) 135

Solution :

By observing the situation practically, in every match there might be three hand shakes.

1  =  Between two players

2  =  Between the winner and umpire

3  =  Between the loser and umpire

So, first let us calculate the number of matches between 10 players.

10 C ₂  =  10!/8!2!

Total number of matches  =  45

Number of hand shakes  =  3(45)

  =  135

So, the required number of hand shakes are 135.

Question 3 :

Find the number of ways in which the letters of the word NUMBER can be scrambled so that the first and the last letters are both vowels

(A) 18           (B)  48           (C) 12

Solution :

Vowels in the given word are U and E.

1   ____   ____   ___   ___   1

We can fix any one of the letters out of U and E. So, the first and last dash may have only one options respectively.

Total number of letters  =  6

We can place any one of the remaining four letters in the second dash, one of the remaining 3 letters in third dash and goes on.

So,

1 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1  =  24

24 + 24  =  48

Hence the total number of ways is 48.

Question 4 :

Two concentric circles have a chord running through the outer one. The chord is the tangent of the inner circle and is 14 cm.The outer circle is shaded and the inner circle is not. Find the exact area of the shaded region without using a calculator.

(A) 35π               (B) 49π            (C) 38π

The area of the outer circle is πR².

The area of the inner circle is πr².

Therefore the area of the shaded region is the difference of the two :

     A  =  π(R² - r²)

Since that is a right-triangle, Pythagorus applies, so

     R² = 7² + r²

     R² - r² = 7²

Thus,

     A  =  π(7²)

 A  =  49π

Question 5 :

Find the area of the circle whose radius is 5 cm.

(A)  25π cm²         (B)  50π cm²        (C)  16π cm²

Solution :

Area of the circle  =  πr²

Radius  =  5

Area of the circle  =  π(5)²

  =  25π

Question 6 :

Let x be an integer that 4, 320, 000 must be multiplied by to get a number with exactly eight terminating zeroes. Find the value of x

(A) 625          (B) 670          (C) 520

Solution :

If we multiply 4,320,000 by 625, we get the answer 2, 700, 000, 000.

It has 8 terminating zeroes.

Question 7 :

What is the area of triangle if the base = 12 cm and height  =  6 cm.

(A) 45 cm²            (B) 25 cm²           (C) 36 cm²

Solution :

Area of the triangle  =  (1/2) ⋅ base ⋅ height

  =  (1/2) ⋅ 12 ⋅ 6

  =  36 cm²

Question 8 :

What is the value of -27+15-16

(A) 28            (B) -28           (C) -58

Solution :

 -27 + 15 - 16

In the given three numbers, two values are having same sign. 

  =  -43 + 15

  =  -28

So, the answer is -28.

Question 9 :

Let p(x) = x², q(x) = 2x, r(x) = p [q(x)] – q [p(x)]. What is the value of r (10)?

(A) 150            (B) 100          (C) 200

Solution :

r(x)  =  p [q(x)] – q [p(x)]

r(x)  =  p(2x) – q(x²)

r(x)  =  (2x)² – 2x²

r(x)  =  4x² – 2x²

r(x)  =  4(10)² – 2(10)²

  =  400 - 200

  =  200

So, the answers is 200.

Question 10 :

What is the value of x which satisfies the below expression x³-x²-x+1 ÷ x³-x²+x-1

(A) 2           (B) 1             (C) -1

Solution :

Let p(x)  =  x³- x²- x + 1 and q(x)  =  x³-x²+x-1

p(2)  ≠  q(2)

p(1)  =  q(1)

So, the answer is 1.

Answer Key

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