EXERCISE 7A (OBJECTIVE TYPE QUESTIONS IN 46 50 SOLUTION )

46.       The sum of the squares of two numbers is 3341 and the difference of their squares s 891. The numbers are:

a)      25,36

b)      25,46

c)       35,46

d)      None of these

47.       The difference between two positive integers is 3. If the sum , of their squares is 369, then the sum of the numbers is:

a)      25

b)      27

c)       33

d)      81

48.       If the sum of two numbers is 22 and the sum of their squares is 404, then the product of the numbers  is:

a)      40

b)      44

c)       80

d)      88

49.       The difference between the squares of two numbers is 256000 and the sum of the numbers is 1000. The numbers are:

a)      600, 400

b)      628,372

c)       640,360

d)      None of these

50.       If the difference of two numbers is 3 and the difference of their squares is 39, then the larger number is:

a)      8

b)      9

c)       12

d)      13

ANSWERS 

46.C

47.B

48.A

49.B

50.A

SOLUTION 

 46.       Let the numbers be X and Y. Then,

X² + Y² = 3341  (I)   and x² - Y²= 891    …(II)

Adding (I) and (II) from (I), we et: 2y²= 2450 or y² = 1225 or y = 35.

So, the numbers are 35 and 46.

47.       Let the nubs box and (X +3). Then,

 X² + (x + 3)² = 369   X² + x² + 9 + 6X = 369

2X² + 6x – 360 = 0      x² + 3x – 180 = 0  (x + 15) (x – 12) = 0   x = 12.

So, the numbers are12 and 15.

48.       Let the numbers be X and y. Then, ( X + Y) = 22 and x ² + Y²= 404.

Now, 2xy = (x + Y)² - (x² + y²) = (22)² - 404 = 484 -404 = 80     xy = 40.

49.       Let the numbers be x and. Then, X² - Y²= 256000 and X + y = 1000.

        On dividing, we get: x – y = 256.

        Solving X + y = 13 and X – y = 256, we get: X = 628 and y = 372.

50.       Let the numbers be X and Y. The, x² - Y² = 39 and x – y = 3.

          On dividing, we get: x+ y = 13.

          Solving x + y= 13 and x – y = 3, we get: x = 8 and y = 5.

          Larger number = 8.