# Number System Word Problems

**Number system word problems**

**Word problems in number system help us form equations easily. This foundation is very important to solve many questions in the upcoming topics. The basic rules of forming equations as follow.**

**1. 'x' exceeds 'y' by 100 ⇒ x – y = 100**

**2. 'x' is twice that of 'y' ⇒ \(\frac{x}{y}\)=2 or x=2y**

**3. 'a' is 25% less than 'b' ⇒ a=(100−25)%b or a=75%(b)**

**4. 'a' is 20% more than 'b' ⇒ a=120%(b)**

**5. \(a^2−b^2\) = (a+b)(a−b)**

**6. \((a+b)^2 = a^2 + 2ab+b^2\)**

**7. \((a−b)^2\) = \(a^2\) - 2ab+\(b^2\)**

**8. (\(a+b)^2\) = \((a−b)^2\)+4ab**

**Solved examples**

**1. \(\frac{5}{6}\) of a number exceeds its \(\frac{3}{4}\) by 10. The number is **

**a. 30**

**b. 60**

**c. 90**

**d. 120**

**Correct Option : D**

**Explanation:**

**Let the number be 'x'.**

**\(\frac{5}{6}\)x−\(\frac{3}{4}\)x=10**

**⇒\(\frac{10x-9x}{12}\)=10**

**⇒x=120**

**2. If 1 is added to the denominator of fraction, the fraction becomes \(\frac{1}{2}\). If 1 is added to the numerator, the fraction becomes \(\frac{2}{3}\). The fraction is **

**a. \(\frac{4}{7}\)**

**b. \(\frac{5}{9}\)**

**c. \(\frac{2}{3}\)**

**d. \(\frac{10}{11}\)**

**Correct Option: B**

**Explanation:**

**Let the required fraction be \(\frac{a}{b}\). Then**

**⇒\(\frac{a}{b+1}\)=\(\frac{1}{2}\)⇒2a–b=1 ---- (1)**

**⇒\(\frac{a+1}{b}\)=\(\frac{2}{3}\) ⇒3a–2b=–3 ------- (2)**

**Solving (1) & (2) we get a = 5, b=9**

**Fraction = \(\frac{a}{b}\)=\(\frac{5}{9}\)**

**3. Five times the first of five consecutive even integers is 4 more than twice the middle. The fourth integer is:**

**A. 8**

**B. 10**

**C. 6**

**D. 4**

**Correct Option: B**

**Explanation:**

**Let the five integers be x, x+2, x+4, x+6, x+8**

**Then, 5x = 4+ 2 \(\times\) (x+4) => 5x = 4 + 2x + 8 => 3x = 12 => x = 4**

**Fourth integer is x+6 = 4+6 = 10**

**4. The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 54. What is the positive difference between the two digits of that number? **

**A. 4**

**B. 5**

**C. 6**

**D. 8**

**Correct Option: C**

**Explanation:**

**Let the ten's digit be x and unit's digit be y.**

**xy value is 10x+y (ex: 23 is expressed as 10 \( \times \) 2 +3)**

**hence, xy - yx = 54**

**(10x+y)-(10y+x) = 54**

**9x-9y=54**

**x-y=6**

**So, positive difference between two digits is 6.**

**5. The difference between a two-digit number and the number obtained by interchanging the digits is 27. What is the positive difference between the sum and the positive difference of the digits of the number if the ratio between the digits of the number is 3 : 2 ?**

**A. 4**

**B. 6**

**C. 8**

**D. 12**

**Correct Option: D**

**Explanation:**

**Let xy is a two digit number.**

**x:y = 3 : 2**

**x=3k, y =2k**

**xy - yx = 27**

**(10x+y) - (10y+x) = 27 => 9x-9y=27 => x-y=3**

**3k-2k=3**

**k=3**

**x = 3 \( \times \) 3, y = 2 \( \times \) 3**

**x = 9, y = 6**

**Positive difference between sum and positive difference of digits of number is (9+6)-(9-6) = 6+6=12**

**6. A two-digit number is such that the product of the digits is 18. When 27 is subtracted from the number, then the digits are reversed. The number is:**

**A. 29**

**B. 92**

**C. 36**

**D. 63**

**Correct Option: D**

**Explanation:**

**Let two digit number is xy.**

**x \( \times \) y =18**

**There are 4 possibilities to get product as 18**

**29, 36, 63, 92**

**xy - 27 =yx**

**xy - yx = 27**

**(10x+y) - (10y+x) = 27**

**9(x-y) = 27**

**x-y=3**

**63 is the only number that satisfies ( Difference of the digits is 3)**

**63-27=36 (verified)**

**Alternate Solution:**

**x \( \times \) y = 18**

**Y = \( \frac{18}{x}\)**

**xy -27 =yx**

**xy - yx =27**

**9(x-y) =27**

**x-y=3**

**x - \( \frac{18}{x}\) =3**

**\(\ x^{2} \) - 18 = 3x**

**\(\ x^{2} \) - 3x - 18 =0**

**By factorisation \(\ x^{2} \) -6x +3x -18 =0**

**x(x-6) + 3(x-6)=0**

**(x+3)(x-6)=0**

**x=6**

**y = \( \frac{18}{x}\) = \( \frac{18}{6}\) = 3**

**xy = 63**

**7. The sum of two numbers is four times their difference. If one of the numbers is 15, the other number is**

**A. 9**

**B. 25 **

**C. 5 \(\frac{1}{3}\) **

**D. A or B**

**Correct Option: D**

**Explanation:**

**(a+b) = 4(a-b) => a+b = 4a - 4b => 3a = 5b**

**Case 1: if a = 15, then b will be 9**

**Case 2: if b = 15, then a will be 25.**

**So, 9 or 25 is the answer.**

**8. The sum of squares of two numbers is 81 and the square of their difference is 49. The product of the two numbers is.**

**A. 15**

**B. 12**

**C. 18**

**D. 16**

**Correct Option: D**

**Explanation:**

**\( a^{2}\) + \( b^{2}\) = 81**

**\((a-b)^{2}\) = 49**

**\( a^{2}\) + \( b^{2}\) - 2ab = 49**

**81 - 2ab = 49**

**81 - 49 = 2ab**

**32 = 2ab**

**16 = ab**

**So, 16 is the product of two numbers.**

**9. If \(\frac{2}{5}\) th of a number decreased by 17 is 15, then the number is **

**a. 30**

**b. 50**

**c. 80**

**d. 90 **

**Correct Option: C**

**Explanation:**

**\(\frac{2}{5}\) x - 17 = 15**

**\(\frac{2}{5}\) x = 32**

**x = 32 \( \times \) \(\frac{5}{2}\) **

**x = 16 \( \times \) 5**

**x = 80**

**10. The difference of two numbers is 17 and the difference of their squares is 51. The sum of two numbers is **

**a. 3**

**b. 4**

**c. 6**

**d. 8**

**Correct Option: A**

**Explanation:**

**Let the numbers be a, b; ⇒ a - b= 17 and \(a^{2}\) - \(b^{2}\) = 51**

**\(\frac{a^{2} - b^{2} }{a - b}\) = a + b**

**\(\frac{51}{17}\) = a + b**

**3 = a + b**

**So, Answer is 3.**

**11. The sum of two numbers is 19 and the difference of their squares is 133. The difference between the number is**

**a. 13**

**b. 5**

**c. 7**

**d. 11**

**Correct Option: C**

**Explanation:**

**Let the numbers be a, b; ⇒ a + b= 19 and \( a^{2}\) - \( b^{2}\) = 133**

**\(\frac{ a^{2} - b^{2} }{a + b}\) = a - b**

**\(\frac{133}{19}\) = a - b**

**7 = a - b**

**So, Answer is 7.**

**12. The sum of the square of three consecutive positive integers is 149. What is the product of two smallest integers?**

**a. 42**

**b. 56**

**c. 48**

**d. 35**

**Correct Option: A**

**Explanation:**

**Let three consecutive integers be x-1, x, x+1**

**\( (x-1)^{2} + (x)^{2} + (x+1)^{2} \) = 149**

**\( x^{2} - 2x+1 + x^{2} + x^{2} + 2x + 1 \) = 149**

**3\(x^{2}\)+2 = 149**

**\( x^{2}\) = 49**

**x = +7 or -7**

**We know three consecutive integers are positive.**

**x = 7**

**So, integers are 7-1, 7, 7+1 => 6, 7, 8**

**Product of two smallest integers is 6 \(\times\) 7 = 42.**

**13. The difference between the squares of two consecutive positive numbers is 29. The numbers are**

**a. 14,15**

**b. 15,16**

**c. 17,18**

**d. 18,19**

**Correct Option: A**

**Explanation:**

**Let the numbers be x and x+1**

**\( (x+1)^{2} \) - \( x^{2} \) = 29**

**\( x^{2} \)+2x+1 - \( x^{2} \) = 29**

**2x + 1 = 29**

**2x = 28**

**x = 14.**

**So, Consecutive numbers are x, x+1 => 14, 15.**

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