Screws weight = 90x100 = 9000gms
bolts weight = 100 x 150 = 15000gms
weight = screws + bolts => 24000 gms =>24 kg
Given entire box weight = 35.5kg
empty box = entire box weight - weight => 35.5kg - 24kg => 11.5kg
so the empty box is 11.5kg.
Last digit for the power of 6 is 6 (always)
Power cycle of 7 is 7, 9, 3, 1.
Now 467/4 gives a remainder of 3
Then the last digit is = 3
Last digit is 6 + 3 = 9.
Given number is 987 = 3 x 7 x 47.
So, required number must be divisible by each one of 3, 7, 47.
None of the numbers in 553681 and 555181 are divisible by 3. While 556581 is not divisible by 7.
Correct answer is 555681.
When we divide 1234 by 8, remainder is 2
When we divide 1235 by 8, remainder is 3
When we divide 1237 by 8, remainder is 5
---> 2 x 3 x 5 = 30
As 30 will not be the remainder because it is greater than 8,
when 30 divided by 8, remainder = 6.
Two digit numbers: The two digits can be 2 and 9: Two possibilities 29 and 92.
Three-digit numbers: The three digits can be 1, 2 and 9 => 3! Or 6 possibilities.
We cannot have three digits as (3, 3, 2) as the digits have to be distinct.
We cannot have numbers with 4 digits or more without repeating the digits.
So, there are totally 8 numbers.
To solve a pair of simultaneous equations such as those given we can add or subtract them.
Adding we get 4x + 4y = 20
Therefore 2x + 2y = 10
Sum of the decimal places = 7 .
Since the last digit to the extreme right wil be zero ( 5 * 4 = 20), so there will be 6 significant digits to the right of the decimal point....
1 to 300 = 300 numbers
squares btwn 1 to 300 = 1,4,9,16,25,36,49.....289 i.e, total 17
cubes = 1,8,27,64,125,216 i.e, 5
now 1 and 64 are common so total numbers that should be removed ( 17+5-2= 20 )
total 20 term removed so 300th term would be 320
A prime number is a whole number greater than 1 whose only factors are 1 and itself.
Factors of 5 are 1, 5
Factors of 11 are 1, 11
Factors of 21 are 1, 3, 7, 21
Factors of 37 are 1, 37.
Hence, according to the definition of a prime number, 21 is not a prime number as it has more than two factors.
1 1
8 A 9
+ 4 C 6
- 6 B 2
7 2 3
We may represent the given sum, as shown below:
1 + A + C - B = 12
A + C - B = 11
Now,giving the maximum values to A and C, i.e.
A = 9 and C = 9, we get B = 7.
Given
Thus, there are (5 + 5 + 9 + 9 + 3)= 31 prime numbers.
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