1.Algebra Booster

8.62 Algebra Booster
20. Out of 2 cards, one is an ace of spade and other card can
take 51 ways.
or 1 ace of any other 3 aces and 1 spade from any other
12 spades in 3 ¥ 12 = 36 ways.
Hence, the required probability,
(51+
36) 87 29 29
= = =
52
C2
26 ¥ 51 26 ¥ 17 442
21. For 7 m + 7 n is be divisible by 5.
Case I: When m ends with 9 and n ends with 1.
Thus, m = 2, 6, 10, …, 98 (25 cases)
and n = 4, 8, 12, …, 100 (25 cases)
Therefore, it can be possible in 25 ¥ 25 ways.
Case II: When m ends with 7 and n ends with 3
Thus m = 1, 5, 9, …, 97 (25 cases)
and n = 1, 5, 9, …, 97 (25 cases)
Therefore, it can also be possible in 25 ¥ 25 ways.
Hence, the required probability
2¥ 2¥ 25¥
25
=
100 ¥ 100
1
=
4
22. Clearly, the last digit of the product of n integers is 1, 2,
3, 4, 6, 7, 8 or 9.
Hence, the required probability
n
Ê 8 ˆ Ê4ˆ
= Á =
Ë
˜ Á ˜
10¯ Ë5¯
23. Hence, the required probability,
365
C =
365 =
1
(365) (365) (365)
1
3 3 2
24. Given equation is 3x = 2x 2 + 1
Clearly, it has three solution at x = 0, 1 and 2.
n
Hence, the required probability = 3 .
10
25. Clearly, 3 integers can be chosen in 20 C 3
ways.
The product of three integers will be even if at least one
of the integers is even.
Hence, the required probability
= 1 – probability that none of three integers is even.
Ê
10
C ˆ
3
= Á1-
20 ˜
Ë C3
¯
10◊9◊8 2 17
= 1- = 1- =
20◊19◊18 19 19
Integer Type Questions
1. For each toss, there are 4 cases will be arise
Case I: A gets head and B gets head
Case II: A gets head and B gets tail
Case III: A gets tail and B gets head
Case IV: Both A and B get tail
So, out of 4 choices, only one choice is there, where A
and B both get tail
50
Ê3ˆ
Hence, the required probability = Á
Ë
˜
4¯
Clearly, p = 3 and q = 4
Hence, the value of (p + q + 2) is 9.
2. We have,
x 2 – 13x £ 30
fi x 2 – 13x – 30 £ 0
fi (x – 3)(x – 10) £ 0
fi 3 £ x £ 10
Thus, x = 3, 4, 5, 6, 7, 8, 9, 10
Hence, the required probability of x = 8 =
4
10 5
Clearly, a = 4 and b = 5.
Hence, the value of (b – a + 2) = 1 + 2 = 3.
3. Hence, the required probability
1 1
¥
4 6 1 1
= = =
3 5 1 1
¥ + ¥
15 + 1 16
4 6 4 6
Clearly m = 1 and n = 16
Ê n ˆ
Hence, the value of Á ˜ is 8.
Ëm
+ 1¯
4. Hence, the required probability,
3 3
C1 C1
6
C2
¥ 3¥ 3¥
2 3
= =
6¥
5 5
Clearly a = 3 and b = 5
Hence, the value of (a + b) is 8.
1 1
5. Clearly PH ( ) = , P(6)
=
2 6
Hence, the required probability of getting a head before
getting 6
2 2 3
1 5Ê1ˆ Ê5ˆ Ê1ˆ
= + Á + + ...
2 6Ë ˜
2¯ Á
Ë
˜
6¯ Á
Ë
˜
2¯
1 1
2 2 6
= = =
5 7
1-
7
12 12
Thus, p = 6/7
Hence, the value of (7p + 1) = 6 + 1 = 7.
6. Here, n(S) = 6 ¥ 6 ¥ 6 ¥ 6 = 1296
Let E be the event of getting a sum 12.
Now, n(E)
n(A) = Co-efficients of x 12 in (x + x 2 + x 3 + … + x 6 ) 4
= Co-efficients of x 8 in (1 + x + x 2 + … + x 5 ) 4
= Co-efficients of
4
6
8
Ê1
- x ˆ
x in Á 1 - x
˜
Ë ¯