111
The value of 
is
is
CTET Solved Paper MCQ - CTET September 2015
We can solve this by using the divisibility rule of 11.
It states that when the sum of digits at odd and even places are equal or differ by a number divisible by 11, then the number is also divisible by 11.
For, 9y8071
Sum of digits at odd places = 9 + 8 + 7 = 24
Sum of digits at even places = y + 0 + 1 = y + 1
⇒ 24 - (y + 1) = 23 - y
∴ To make this number divisible by 11, we have to put the value of y as 1.
It states that when the sum of digits at odd and even places are equal or differ by a number divisible by 11, then the number is also divisible by 11.
For, 9y8071
Sum of digits at odd places = 9 + 8 + 7 = 24
Sum of digits at even places = y + 0 + 1 = y + 1
⇒ 24 - (y + 1) = 23 - y
∴ To make this number divisible by 11, we have to put the value of y as 1.
(14.96) × 1010 - (3.84) × 108
= (1496) × 108 - (3.84) × 108
= (1496 - 3.84) × 108
= 1492.16 × 108
= 1.49216 × 1011
= (1496) × 108 - (3.84) × 108
= (1496 - 3.84) × 108
= 1492.16 × 108
= 1.49216 × 1011
6, 3, 9, 5, 1, 2, 3, 6, 5, 1, 3
Arranging them in ascending order, we get
1, 1, 2, 3, 3, 3, 5, 5, 6, 6, 9
Mode of the data = 3, Median of the data = 3
∴ Sum of mean, mode, median = 4+3+3 = 10
Arranging them in ascending order, we get
1, 1, 2, 3, 3, 3, 5, 5, 6, 6, 9
Mode of the data = 3, Median of the data = 3
∴ Sum of mean, mode, median = 4+3+3 = 10
Total number of outcomes = 12
Favourable number of outcomes (to be a prime number) = 4 (i.e 5, 7, 11, 13)
Favourable number of outcomes (to be a prime number) = 4 (i.e 5, 7, 11, 13)
