EXERCISE 7A (OBJECTIVE TYPE QUESTIONS IN 41 45 SOLUTION )

41.       The sum, of two numbers is 40 and their product is 375. What will be sum of their reciprocals?

a)      1/40

b)      8/75

c)       75/4

d)      75/8

42.       The sum of two positive integers multiplied by the bigger number is 204, and the difference multiplied by the smaller umber is 35. The numbers are:

a)      12 ,5

b)      13,4

c)       14,3

d)      24,10

43.       If the sum and difference of two numbers are 20 and 8 respectively, then the difference of their squares is:

a)      12

b)      28

c)       160

d)      180

44.       The product of two numbers is 120 and the sum of their squares is 289. The sum of the numbers is :

a)      20

b)      28

c)       160

d)      180

45.       The product of two numbers is 45 an the sum of their squares is 106. The numbers are:

a)      3 and 5

b)      5 an 9

c)       5 an 19

d)      45 an 1

ANSWERS 

41.B

42.A

43.C

44.B

45.B

SOLUTION 

41.       Let the numbers be X and y. Then, X + Y = 40 and xy = 375.

1/x + 1/y = x + y / xy = 40 / 375 = 8 / 75.

42.       Let the numbers be X and y such than x > y. Then,

X (X + Y ) = 204       X ² + x y = 204  …(I) an y (x – y) = 35   xy = y² = 35 ..(II)

Subtracting (II) from (I), we get: X ² + Y² = 169.

The only triplet satisfying this condition is (12, 5, 13,) Thus, X = 12, Y = 5.

43.       Let the numbers be X and Y. then, x + y = 20 and x – y = 8.

 X²- Y² = (x + Y ) (x – Y ) = 20 x 8= 160.

44.       Let the numbers be an defy. Then, xy = 120 an x² + Y² = 289.

(X + Y)² =x² + y² + 2xy = 289 + 240 = 529.

X +y = 529 = 23.

45.       Let the numbers be x and y. Then, xy = 45 and X²  + Y² = 106.

( X + y) = (x² + Y²) + 2xy= 106 + 90 = 196      x + y = 14.      …..(I)

( X – Y ) = (( X² + Y²) – 2 xy= 16   X – y = 4  …. (II)

Solving (I) and (II), we get: x = 9 and y = 5.