Problems on Numbers Questions and Solutions

• 8. Product of two natural numbers is 17. Then, the sum of reciprocals of their squares is

  1. \begin{aligned} \frac{290}{289} \end{aligned}
  2. \begin{aligned} \frac{1}{289} \end{aligned}
  3. \begin{aligned} \frac{290}{90} \end{aligned}
  4. \begin{aligned} \frac{290}{19} \end{aligned}

  Answer :

  Option A

  Explanation:

  If the numbers are a, b, then ab = 17,
  as 17 is a prime number, so a = 1, b = 17.
  
  \begin{aligned} \frac{1}{a^2} + \frac{1}{b^2} =
  \frac{1}{1^2} + \frac{1}{17^2}
  \end{aligned}
  \begin{aligned} = \frac{290}{289}
  \end{aligned}

• 9. Two numbers differ by 5. If their product is 336, then sum of two number is

  1. 33
  2. 34
  3. 36
  4. 37

  Answer :

  Option D

  Explanation:

  Friends you remember,
  \begin{aligned}
  => (x+y)^2 = (x-y)^2 + 4xy
  \end{aligned}
  
  \begin{aligned}
  => (x+y)^2 = (5)^2 + 4(336)
  \end{aligned}
  
  \begin{aligned}
  => (x+y) = \sqrt{1369} = 37
  \end{aligned}

• 10. Sum of three numbers 264, If the first number be twice then second and third number be one third of the first, then the second number is

  1. 70
  2. 71
  3. 72
  4. 73

  Answer :

  Option C

  Explanation:

  Let the second number is x, then first is 2x, and third is 1/3(2x)
  \begin{aligned}
  =>2x + x + \frac{2x}{3} = 264 <=> \frac{11x}{3} = 264
  \end{aligned}
  
  \begin{aligned}
  => x = 72
  \end{aligned}

• 11. Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is

  1. 12
  2. 13
  3. 15
  4. 17

  Answer :

  Option C

  Explanation:

  Let the three integers be x, x+2 and x+4.
  Then, 3x = 2(x+4)+3,
  x= 11
  Therefore, third integer x+4 = 15

• 12. Sum of two numbers is 25 and their difference is 13. Find their product.

  1. 104
  2. 108
  3. 114
  4. 124

  Answer :

  Option C

  Explanation:

  Friends, this sort of question is quite important in competitive exams, whenever any question come which have relation between sum, product and difference, this formula do the magic:
  \begin{aligned}
  => (x+y)^2 = (x-y)^2 + 4xy
  \end{aligned}
  
  \begin{aligned}
  <=> (25)^2 = (13)^2 + 4xy
  \end{aligned}
  
  \begin{aligned}
  <=> 4xy = (25)^2 - (13)^2
  \end{aligned}
  \begin{aligned}
  <=> xy = \frac{456}{4} = 114
  \end{aligned}

• 13. find the number, If 50 is subtracted from two-third of number, the result is equal to sum of 40 and one-fourth of that number.

  1. 214
  2. 216
  3. 114
  4. 116

  Answer :

  Option B

  Explanation:

  Let the number is x,
  \begin{aligned}
  => \frac{2}{3}x-50 = \frac{1}{4}x + 40
  \end{aligned}
  
  \begin{aligned}
  <=> \frac{2}{3}x-\frac{1}{4}x = 90
  \end{aligned}
  \begin{aligned}
  <=> \frac{5x}{12} = 90
  \end{aligned}
  \begin{aligned}
  <=> x = 216
  \end{aligned}

• 14. Sum of two numbers is 40 and their difference is 4. The ratio of the numbers is

  1. 10:3
  2. 5:9
  3. 11:9
  4. 13:9

  Answer :

  Option C

  Explanation:

  \begin{aligned}
  => \frac{(x+y)}{(x-y)} = \frac{40}{4}
  \end{aligned}
  
  \begin{aligned}
  => (x+y)= 10(x-y)
  \end{aligned}
  
  \begin{aligned}
  => 9x = 11y => \frac{x}{y} = \frac{11}{9}
  \end{aligned}