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
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Q2. 

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
=
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Q3. 

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 = A_{3}= a+2d =13 (i)
8th term = A_{8}= a+7d =2  (ii)
Subtracting equation (i) from (ii), we get :
or 7d2d=2(13)
or 5d =15
or d=15/5=3
Substituting it in equation (ii),a=27(3)=221=19
Therefore, 14th term =
A_{14}= a + 13d = 19 + 13(3) = 19 + 39 = 20
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Q4. 

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 Answer
Q5. 

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 Answer
Q6. 

1)  53  2)  57 
3)  63  4)  72 
5)  None of these 
Answer : 53
Explanation :
Explanation :
Given,T_{3} = a + 2d = 13(1)
T_{5} = a + 4d = 21(2)
on solving (1) and (2)
d = 4 & a = 5
T _{13}= a + 12d = 5 + 12(4) = 5 + 48 = 53
View Answer
Q7. 

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 Answer
Q8. 

1)  14  2)  16 
3)  17  4)  20 
5)  None of these 
Answer : 16
Explanation :
Explanation :
Given,a_{1}= 6 and d =126 = 6
Since S_{ n}=816
Sum of n terms of A.P. is given as, S_{n}=n/2[2a+(n1)d]
816= n/2[2x6+(n1)6]
or 1632= n[12+6n6] or 1632= n[6n+6]
or 1632= 6n2 + 6n or 6n^{2} + 6n 1632 = 0
Divide both sides by 6, we get,n^{2} + n272= 0 or n^{2} + 17n 16n 272 = 0
or n(n+17)  16(n+17) = 0 or n = 16
View Answer
Q9. 

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
11a7a = 42d110d
4a = 68d a = 17d
A_{18 }= a+17d = 17d+ 17d = 0
View Answer
Q10. 

1)  18  2)  19 
3)  21  4)  24 
5)  None of these 
Answer : 21
Explanation :
Explanation :
T_{7 }= a + 6d = 15 ....... (1)
T_{12} = a + 11d = 5 ...... (2)
On solving (1) and (2), d = 4 & a = 39
T_{16} = a + 15d = 39 + 15(4) = 39 + 60 = 21
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