Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q531. |
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| 1) | 1/2 | 2) | 1/3 |
| 3) | 3/4 | 4) | 4/5 |
| 5) | None of thesae | ||
Answer : 3/4
Explanation :
Explanation :
In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.
Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1),
(4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(E) = 27.
So, P(E) = n(E) / n(S) =27 /36 = 3/4
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Q532. |
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| 1) | 1/24 | 2) | 21/46 |
| 3) | 1/48 | 4) | 24/117 |
| 5) | None of these | ||
Answer : 21/46
Explanation :
Explanation :
Let S be the sample space and E be the event of selecting 1 girl and 2 boys.
Then, n(S) = 25C3 = Number ways of selecting 3 students out of 25
= (25 x 24 x 23) / (3 x 2 x 1) = 2300.
n(E) = (10C1 x 15C2)
= 10 x (15 x 14) /(2 x 1) = 1050.
So, P(E) = n(E) / n(S) =1050 / 2300 =21/46
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Q533. |
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| 1) | 1/4 | 2) | 3/4 |
| 3) | 7/8 | 4) | 1/36 |
| 5) | None of these | ||
Answer : 7/8
Explanation :
Explanation :
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = Event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
So, P(E) = n(E) / n(S) =7 /8 .
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Q534. |
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| 1) | 1/6 | 2) | 1/8 |
| 3) | 1/9 | 4) | 1/12 |
| 5) | None of these | ||
Answer : 1/9
Explanation :
Explanation :
In two throws of a die, n(S) = (6 x 6) = 36.
Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.
So,P(E) = n(E) / n(S) = 4 /36 =1/9.
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Q535. |
|
| 1) | 60 | 2) | 120 |
| 3) | 180 | 4) | 240 |
| 5) | None of these | ||
Answer : 120
Explanation :
Explanation :
The word 'DELHI' has 5 letters and all these letters are different.
Total words (with or without meaning) formed by using all these 5 letters using each letter exactly once
= Number of arrangements of 5 letters taken all at a time = 5P5 = 5! = 5 x 4 x 3 x 2 x 1 = 120
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Q536. |
|
| 1) | 60 | 2) | 120 |
| 3) | 240 | 4) | 360 |
| 5) | None of these | ||
Answer : 360
Explanation :
Explanation :
The word 'LEADER' has 6 letters.
But in these 6 letters, 'E' occurs 2 times and rest of the letters are different.
Hence,number of ways to arrange these letters = 6!2!=6×5×4×3×2×12×1 = 360
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Q537. |
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| 1) | 8750 | 2) | 10420 |
| 3) | 11760 | 4) | 12420 |
| 5) | None of these | ||
Answer : 11760
Explanation :
Explanation :
We need to select 5 men from 8 men and 6 women from 10 women
Number of ways to do this = 8C5 x 10C6
= 8C3 x 10C4 [Applied the formula nCr = nC(n - r) ]
= (8×7×63×2×1) (10×9×8×74×3×2×1)
= 56 x 210 = 11760
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Q538. |
|
| 1) | 2100 | 2) | 24400 |
| 3) | 21300 | 4) | 25200 |
| 5) | None of these | ||
Answer : 25200
Explanation :
Explanation :
Number of ways of selecting 3 consonants out of 7 = 7C3
Number of ways of selecting 2 vowels out of 4 = 4C2
Number of ways of selecting 3 consonants out of 7 and 2
vowels out of 4 = 7C3 x 4C2
=(7×6×5/3×2×1)×(4×3/2×1)=210
It means that we can have 210 groups where each group
contains total 5 letters(3 consonants and 2 vowels).
Number of ways of arranging 5 letters among themselves = 5! = 5 x 4 x 3 x 2 x 1 = 120
Hence, Required number of ways = 210 x 120 = 25200
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Q539. |
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| 1) | 100800 | 2) | 120960 |
| 3) | 240150 | 4) | 4989600 |
| 5) | None of thesae | ||
Answer : 120960
Explanation :
Explanation :
In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging these letters = 8! / (2!)(2!)= 10080.
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = 4! /2! = 12.
So, required number of words = (10080 x 12) = 120960.
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Q540. |
|
| 1) | 42 | 2) | 48 |
| 3) | 63 | 4) | 90 |
| 5) | None of these | ||
Answer : 63
Explanation : The required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = (7 x 6 )/(2 x 1) x 3 = 63.
Explanation : The required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = (7 x 6 )/(2 x 1) x 3 = 63.
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