Number System - Study Mode

[#211] The sum of four consecutive two-digit odd numbers, when divided by 10, become a perfect square. Which of the following can possibly be one of these four numbers?
Correct Answer

(C) 41

Explanation

Solution: Using options, We find that four consecutive odd numbers are 37, 39, 41 and 43 The sum of these 4 numbers is 160, when divided by 10 we get 16 which is a perfect square. Thus, 41 is one of the odd numbers

[#212] Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?
Correct Answer

(D) 1

Explanation

Solution: Any four digit number in which first two digits are equal and last two digits are also equal will be in the form 11 × (11a + b) i.e. it will be the multiple of 11 like 1122, 3366, 2244, . . . . Now, let the required number be aabb. Since aabb is a perfect square, the only pair of a and b that satisfy the above mentioned condition is a = 7 and b = 4
Hence, 7744 is a perfect square

[#213] On a road three consecutive traffic lights change after 36, 42 and 72 seconds respectively. If the lights are first switched on at 9:00 AM sharp, at what time will they change simultaneously?
Correct Answer

(B) 9:08:24

Explanation

Solution: LCM of 36, 42 and 72, 36 = 2 × 2 × 3 × 3 42 = 2 × 3 × 7 72 = 2 × 2 × 2 × 3 × 3 LCM = 2 ×2 × 2 × 3 × 3 × 7 = 504 seconds. LCM of 36, 42 and 72 is 504 Hence, the lights will change simultaneously after 8 minutes and 24 seconds.

[#214] The HCF of 2472, 1284 and a third number 'N' is 12. If their LCM is 2 3 × 3 2 × 5 × 103 × 107, then the number 'N' is:
Correct Answer

(D) None of these

Explanation

Solution: We have, HCF of the numbers × LCM of the numbers = Multiplication of the numbers. (12) × (2 3 × 3 2 × 5 × 103 × 107) = 2472 × 1284 × N Hence, N = $$frac{{{ left( {12}
ight) imes left( {{2^3} imes {3^2} imes 5 imes 103 imes 107}
ight)} }}{{2472 imes 1284}}$$ Or, N = 3 × 5

[#215] Find the LCM and HCF of 2.5, 0.5 and 0.175.
Correct Answer

(D) 17.5

Explanation

Solution: $$eqalign{
& 2.5 = frac{{25}}{{10}}, cr
& 0.5 = frac{5}{{10}}, cr
& 0.175 = frac{{175}}{{1000}}, cr} $$ Now, LCM of two or more fractions is given by: $$eqalign{
& frac{{{ ext{LCM}},{ ext{of}},{ ext{Numerators}}}}{{{ ext{HCF}},{ ext{of}},{ ext{Denominators}}}} cr
& { ext{Thus,}} cr
& frac{{{ ext{LCM}},{ ext{of}},25,,5,,175}}{{{ ext{HCF}},{ ext{of}},10,,10,,1000}} cr
& = frac{{175}}{{10}} cr
& = 17.5 cr} $$