Section 'B'
2
Ouestion numbers 21 to 26 carry 2 marks each.
The HCF and LCM of two numbers are 9 and 360 respectively. If one number is 45, find the other
number
OR
Show that 7-15 is irrational, given that V5 is irrational.
22. Find the 20th term from the last term of the AP 3, 8, 13,.....,253.
OR
If 7 times the 7th term of an A.P is equal to 11 times its 11th term, then find its 18th term.
23. Find the coordinates of the point P which divides the line joining of A (-2,5) and B (3,-5) in the ratio
2:3.
24. A card is drawn at random from a well shuffled deck of 52 cards. Find the probability of getting neither
a red card nor a queen.
25. Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find
the probability that the product is a prime number.
26. For what value of p will the following pair of linear equations have infinitely many solutions.
(p-3x + 3y =p
px + py = 12
2
NN
Section 'C'
Question numbers 27 to 34 carry 3 marks each.
27. Use Euclid's Division Algorithm to find the HCF of 726 and 275.
3
Answers
H.C.F x L.C.M = Product of numbers
9 x 360 = 45 x Second number
9x360 /45 = second number
second number = 1 x 360 / 5
Second number = 72
Therefore the other number is 72 .
Answer:
HCF×LCM=Product of two number
9×360=45×x
x=3240/45
x=72
answer 22) :
the 20th term from the last term of the AP 3, 8, 13,.....,253.is
an =l-(n-1) d
=253-(20-1) ×5
=253-(19) 5
=253-95
=158
Here p point divide A B in the ratio 2:3 so applying section formula we get
2×3 +3×(-2) /2 +3 and 2×(-5) + 3×5 /2+3
6-6/5 and -10 +15/5
0 and 5/5
0 and 1
hence the cordinates of point P(0,1)
solution
probability
n(S) = 52
Event = {getting neither a red card nor a queen}
∴ There are 26 red cards and 2 more queens are there.
Number of cards each one of which is either a red card or a queen = 28
The event that the card drawn is neither a red card nor a queen = 52 – 28 = 24.
n(E) = 24
n(S) = 52
P(E) = ?
∴ P(E) = n(E)/n(S) = 24/52 = 6/13
Given: two dice are thrown at the same time and the product of numbers appearing on them is noted
To find: the probability that the product is a prime number
Explanation: Total number of outcomes of one dice = 6
So total number of outcomes = 6 × 6 = 36
So, n(S) = 36
And all the outcomes of S are
S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
Let E denote outcomes such that the product is prime (i.e., having only factors), then the favourable outcomes of event E are
E = {(1,2), (1,3), (1,5), (2,1), (3,1), (5,1)}
Hence number of outcomes of E are n(E) = 6
P (getting the product is a prime number)
n(E) /n(S)
Substituting corresponding values, we get
= 6/36
1/6
=
Hence the probability that the product is a prime number is 1/6
p = 6 will make the given pair of linear equations have infinitely many solutions. Since, p = 6 satisfies both equation. Thus, p = 6 will make the given pair of linear equations have infinitely many solutions.
using Euclid’s Division Algorithm, we find HCF of two positive numbers by repetitive division till we get 0 as the remainder.
In the given two numbers 726 and 275, 726 is greater, so we will divide 726 by 275, we get,
We see that the remainder is 176≠0,
Now we divide 275 by 176, we get
We see that the remainder is 99≠0,
Now we divide 176 by 99, we get
We see that the remainder is 77≠0
Now we divide 99 by 77, we get,
We see that the remainder is 22≠0
Now we divide 77 by 22, we get,
We see that the remainder is 11≠0,
Now we divide 22 by 11, we get
We see that the remainder is 0.
Hence the HCF of the two numbers 726 and 275 is 11.
