Let the numbers be x and y. Then, x + y = 20 and x - y = 8.
x^2 - y^2 = (x + y) (x - y) = 20 x 8 = 160.
Let the smallest number be x.
Then larger number = (x + 1365)
x + 1365 = 6x + 15
= 5x = 1350
x = 270
Smaller number = 270.
LCM of 4, 6, 8 and 10 = 120
120) 1000 (8
960
------
40
The least number of four digits which is divisible by 4, 6, 8 and 10 => 1000 + 120 - 40 = 1080.
First line will cut all other 14, similarly second will cut 13, and so on
Total = 14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 105.
Prime Numbers :: Numbers which are divisible by only 1 and itself are Prime Numbers.
It's answer will be 91.
Because 91 can be divisible by 7,13,91,1.
It is quite clear that prime number should be divisible only by itself and by 1.
Given
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 x 0 + 1 + 1
Using BODMAS Rule,
As multiplication precedes addition, 1 x 0 = 0,
Now, 10 + 0 + 1 + 1 = 10 + 2 = 12.
Hence, 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 x 0 + 1 + 1 = 12.
As per given
7th stop only one was left and in each preceding stop twice the nos... there were so 6 stops before 7th.
so
Let the five consecutive odd numbers be x-4, x-2, x, x+2, x+4
According to the question,
Difference between square of the average of first two odd number and the of the average last
two odd numbers is 396
i.e, x+3 and x-3
Hence, the smallest odd number is 33 - 4 = 29.
He sold 4 cycles for Rs. 8400
[It means each cycle price is Rs. 2100]
And again he sold 3 cycles for Rs. 1900 each
If he want to gain avg profit on each cycle is Rs. 320 per cycle.
So profit for 10 cycles is @ Rs.320 is,
10 x 320 = Rs. 3200
He already got Rs. 1850 profit by selling 7 cycles.
Now, remaining amount is 3200 - 1850 = 1350
So from remaining 3 cycles we want to get Rs. 1350 profit ,
1350/3 = 450
so he must sell remaining 3 cycles at Rs. 2200 each one.
Total number of exams written by 100 students = 48 + 45 + 38 = 131
Now let us say x members are writing only 1 exam, y members are writing only 2 exams, z members are writing only 3 exams.
Therefore, x + 2y + 3z = 131 also x + y + z = 100.
Given that z = 5. So x + 2y = 116 and x + y = 95.
Solving we get y = 21.
So 21 members are writing exactly 2 exams.
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