Number System - Study Mode
[#191] Which of the following fraction is greater than $$frac{3}{4}$$ but less than $$frac{5}{6}$$ ?
Correct Answer
(C) $$frac{4}{5}$$
Explanation
Solution: $$eqalign{
& frac{3}{4} = 75\% ,frac{5}{6} = 83.33\% , cr
& (1)frac{2}{3} = 66.66\% , cr
& (2)frac{1}{2} = 50\% , cr
& (3)frac{4}{5} = 80\% , cr
& (4)frac{9}{{10}} = 90\% cr
& { ext{Hence , }}frac{4}{5}{ ext{ is between the fraction}}{ ext{.}} cr} $$
[#192] The digit in unit's place of the product 81 × 82 × 83 × ..... × 89 is :
Correct Answer
(A) 0
Explanation
Solution: 81 × 82 × 83 ×..... × 89 take unit digit multiply 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 = 0 Since there is 5 and 2 are present So we get 5 × 2 = 10 and the last digit of 10 is 0, so the last digit of the above number is also 0
[#193] $$left( {999frac{{999}}{{1000}} imes 7}
ight)$$
xa0 is equal to -
Correct Answer
(D) $${ ext{6999}}frac{{993}}{{1000}}$$
Explanation
Solution: $$eqalign{
& = left( {999frac{{999}}{{1000}} imes 7}
ight) cr
& = left( {999frac{{999}}{{1000}}}
ight) imes 7 cr
& = 6993 + frac{{6993}}{{1000}} cr
& = 6993 + 6frac{{993}}{{1000}} cr
& = 6993 + 6 + frac{{993}}{{1000}} cr
& = 6999frac{{993}}{{1000}} cr} $$
[#194] Number 2, 4, 6, 8, 10.....196, 198, 200 are multiplied together. The number of zeros at the end of the product on the right will be equal to -
Correct Answer
(C) 24
Explanation
Solution: $$eqalign{
& { ext{2, 4, 6, 8, 10}}......{ ext{198, 200 }} cr
& {2^{100}}(1 imes 2 imes 3 imes ..... imes 99 imes 100) cr} $$ $$eqalign{
& underline {5,,,|,,100} cr
& underline {5,,,|,,,20} ,,,,, = 24{ ext{ zero's (20 + 4)}} cr
& ,,,,,|,,,4 cr} $$ When we multiply the series of 1 × 2 × 3 ×..... × 100 then we find 24 zero at the end of the product
[#195] The product of two positive integers is 2048 and one of them is twice the other. Then the small of the number is :
Correct Answer
(A) 32
Explanation
Solution: Let first number is x and second number is 2x. Then, according to the question x × 2x = 2048 x 2 = 1024 x = 32 Smaller number = 32