EXERCISE 7.. PROBLEMS ON NUMBERS(SOLVED EXAMPLES IN 11 14)

EX 11. The ratio between a two – digit number and the sum of the digits of that number is 4:1. If the digit in the unit’s place is 3 more than the digit in the ten’s place what is the number?

Sol. let the ten’s digit be x. then, unit’s digit = (X + 3).

Sum of the digits = X + (X + 3) = 2X + 3. Number = 10x + (X + 3) = 11x + 3.

11X + 3 / 2x + 3 = 4/1    11X + 3 = 4 (2X + 3)   3X + 9   X = 3.

Hence, required number = 11X + 3 = 36.

Ex. 12. A number consists of two digits. The sum of the digits is 9. If 63 is subtracted from the number, its digits are interchanged. Find the number.

Sol. let the ten’s digit be. Then, unit’s digit = (9 –X).

Number = 10X + (9-x) = 9X + 9.

Number obtained by reversing the digits = 10 (9-x) + = 90 – 9x.

(9X + 9) – 63 = 90 – 9X  18X = 144    X = 8.

So, ten’s digit = 8 and unit’s digit = 1.

Hence, the required number is 81.

Ex. 13. A fraction becomes 2/3 when 1 is added to both, its numerator and denominator. Ned, it becomes ½ when 1 is subtracted from both the numerator and denominator. Find the fraction.

Sol. let the required fraction be x/ y. then,

X +1 / Y + 1 = 2/3   3X – 2y = -1 …(I) and X – 1/ Y -1     2X – y = 1…(II)

Solving (I) and (II), we get: X = 3, Y = 5.

Required fraction = 3/5.

Ex.14. 50 is divided into two parts such that the sum of their reciprocals is 1/12. Find the two parts.

Sol. let the two parts be x and (50 –x)

Then, 1/x + 1/50 –X = 1/12  50 –X +X / X (50 –X) = 1/12   x² - 50X + 600 = 0.

(X – 30) (X -20) =0        X = 30 or X = 20.

So, the parts are 30 and 20.

Ex. 15. If three numbers are added in pairs, the sums equal 10, 19 and21. Find the numbers.

Sol. let the numbers be x, y and z. then,

X + y = 10 ..(I)   Y + z = 19 …(II)  X + z = 21 …(III)

Adding (I), (II) and (III), we get: 2(X + Y + Z) = 50 or (X + Y + z) = 50 or (X + Y + z) = 25.

Thus, X = (25 – 19) =6; y = (25 – 21) =4; Z = (25 – 10) = 15.

Hence, the required numbers are 6, 4 and 15.