EXERCISE 7 A (OBJECTIVE TYPE QUESTIONS IN 76 80 SOLUTION )

76.       The digit in the unit’s place of a number is equal to the digit in the ten’s place of half of that number and the digit in the ten’s place of that number is less than the digit in unit’s place of half of the number by 1. If the sum of the digits of the number is7, then what is the number?

a)      34

b)      52

c)       162

d)      Data inadequate

e)      None of these

77.       In a two – digit number, the digit in the unit’s place is more than two vice the digit in ten’s place by. If the digits in the unit’s place and the ten’s place are interchanged, difference between the newly formed number and the original number is less than the original number by 1. What is the original number?

a)      25

b)      37

c)       49

d)      52

e)      73

78.       A certain number of two digits is three times the sum of its digits and if 45 be added to it, the digits are revered. The number is:

a)      23

b)      27

c)       32

d)      72

79.       A two – digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is:

a)      18

b)      24

c)       42

d)      81

80.       The product of two fractions is 14/15 and their quotient is 35/24. The greater fraction is:

a)      4/5

b)      7/6

c)        7/4

d)      7/5

ANSWERS

76. B

77.B

78. B

79. B

80. B

SOLUTION 

76.       Let the ten’s digit be X and unit’s digit be Y. Then,10X + y / 2 = 10 y + (X + 1)

                 10 X + Y = 20Y + 2x+ 2     8x -19Y = 2   …….(I)    and X +y = 7 …..(III)

                  Solving, (I) and (II), we get: X = 5, y = 2. Hence, required number = 52.

77.       Let the ten’s digit be x. Then, Unit’s digit = 2x +1.

                  [ 10X + (2X + 1) ] – [{10 (2X + 1) +X} – (10X + (2X + 1}] = 1

                 (12X +1) – (9X + 9) = 1    3X = 9      X = 3.

                 So, ten’s digit = 3 and unit’s digit =7. Hence, Original number = 37.

78.       Let the ten’s digit be X and unit’s digit be y.

                   Then, 10X + y = 3 (X + Y)       7x – 2y = 0

                   1X+y + 45 = 10y +X      y – X = 5

                   Solving (I) and (II), we get: X = 2 and Y =7.

                   Required number = 27.

79.       Let the ten’s and unit’s digit be X and 8/x respectively.

                   Then, (10X + 8 / X ) + 18   = 10 x 8 ? X + X    10x² + 8 X 18X = 80 + x²

                    9X² + 18x -72 = 0  x² + 2x – 8 = 0     (X + 4) (X – 2) = 0  x = 2.

                    So, ten’s digit = 2 and unit’s digit = 4. Hence, required number = 24.

80.       Let the two fractions be A  and B. then, ab = 14 / 15 and 1/b = 35 / 24.

                   Ab / (a / b) = (15 / 15 x 24 / 35 )      b² = 16 / 25      b = 4/5. a b = 14 / 15    a = (14 / 15                    x 5/4 )  = 7 / 6.

                   Since a>b, so greater fraction is 7/6.