Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q1. |
|
| 1) | 8 | 2) | 10 |
| 3) | 11 | 4) | 14 |
| 5) | None of these | ||
Answer : 11
Explanation :
Explanation :
Explanation :
Let, the number of subjects = x
Then, Total marks he scored for all subjects = 63x
If he had obtained 20 more marks for his Physics and 2 more marks for his Chemistry, his average would have been 65.
=> Total marks he would have scored for all subjects = 65x
Now, we can form the equation as 65x - 63x = the additional marks of the student
= 20 + 2 = 22
=> 2x = 22
=>x=22/2=11
View AnswerQ2. |
|
| 1) | 5/12 | 2) | 7/12 |
| 3) | 11/12 | 4) | 3/11 |
| 5) | None of these | ||
Answer : 7/12
Explanation :
Explanation :
Total number of trials = 300 (Since there are 300 students all together).
Number of times a cricket player is chosen = 95 (Since 95 students play cricket).
Number of times a football player is chosen = 120.
Number of times a volleyball player is chosen = 80.
Number of times a student is chosen who plays no games = 5.
The probability of getting a player who plays either cricket or volleyball
= (95+80)/300 = 175/300 = 7/12
=
View AnswerQ3. |
|
| 1) | 18 | 2) | 20 |
| 3) | 28 | 4) | 32 |
| 5) | None of these | ||
Answer : 20
Explanation :
Explanation :
Let,the first term of an AP = a and the common difference = d
3th term of AP = A3= a+2d =-13 -------(i)
8th term = A8= a+7d =2 ------- (ii)
Subtracting equation (i) from (ii), we get :
or 7d-2d=2-(-13)
or 5d =15
or d=15/5=3
Substituting it in equation (ii),a=2-7(3)=2-21=-19
Therefore, 14th term =
A14= a + 13d = -19 + 13(3) = -19 + 39 = 20
View AnswerQ4. |
|
| 1) | 84 | 2) | 102 |
| 3) | 106 | 4) | 108 |
| 5) | None of these | ||
Answer : 102
Explanation :
Explanation :
First term of AP = a=-19 and last term = l=36
Number of terms = n=12
Sum of AP = n/2(a+l)
= 12/2(-19+36) = 17x6=102
View AnswerQ5. |
|
| 1) | 68 | 2) | 70 |
| 3) | 72 | 4) | 76 |
| 5) | None of these | ||
Answer : 68
Explanation :
Explanation :
Given, First term of AP = a =- 20 and last term = l = 28
Number of terms = n = 17
Sum of AP = n/2(a+l)= 17/2(-20+28)= 17x4 = 68
View AnswerQ6. |
|
| 1) | 53 | 2) | 57 |
| 3) | 63 | 4) | 72 |
| 5) | None of these | ||
Answer : 53
Explanation :
Explanation :
Given,T3 = a + 2d = 13-------(1)
T5 = a + 4d = 21-------(2)
on solving (1) and (2)
d = 4 & a = 5
T 13= a + 12d = 5 + 12(4) = 5 + 48 = 53
View AnswerQ7. |
|
| 1) | 11 | 2) | 13 |
| 3) | -11 | 4) | -13 |
| 5) | None of these | ||
Answer : 11
Explanation : Given, the 7th and the 10th terms are (a + 6d) and (a + 9d) respectively.
Explanation : Given, the 7th and the 10th terms are (a + 6d) and (a + 9d) respectively.
The ratio of these terms is -1.
Solving this ratio, we get -2a = 15d. The 16th term is (a + 15d) = -15.
Substituting for a, we get a = 15 and d = -2. Therefore, the 3rd term is (a + 2d) = 11.
View AnswerQ8. |
|
| 1) | 14 | 2) | 16 |
| 3) | 17 | 4) | 20 |
| 5) | None of these | ||
Answer : 16
Explanation :
Explanation :
Given,a1= 6 and d =12-6 = 6
Since S n=816
Sum of n terms of A.P. is given as, Sn=n/2[2a+(n-1)d]
816= n/2[2x6+(n-1)6]
or 1632= n[12+6n-6] or 1632= n[6n+6]
or 1632= 6n2 + 6n or 6n2 + 6n -1632 = 0
Divide both sides by 6, we get,n2 + n-272= 0 or n2 + 17n -16n -272 = 0
or n(n+17) - 16(n+17) = 0 or n = 16
View AnswerQ9. |
|
| 1) | 1 | 2) | 0 |
| 3) | 12 | 4) | 18 |
| 5) | None of these | ||
Answer : 0
Explanation :
Explanation :
Let, the first term of the AP be 'a' and the common difference be 'd'.
7th term = A 7= a + 6d
11th term = A 11 = a + 10d
A.T.Q., 7(a+6d)=11(a+10d) = 7a + 42d = 11a + 110d
11a-7a = 42d-110d
4a = -68d a = -17d
A18 = a+17d = -17d+ 17d = 0
View AnswerQ10. |
|
| 1) | 18 | 2) | 19 |
| 3) | 21 | 4) | 24 |
| 5) | None of these | ||
Answer : 21
Explanation :
Explanation :
T7 = a + 6d = -15 ....... (1)
T12 = a + 11d = 5 ...... (2)
On solving (1) and (2), d = 4 & a = -39
T16 = a + 15d = -39 + 15(4) = -39 + 60 = 21
View Answer