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ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams

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##### In an examination, a student's average marks were 63. If he had obtained 20 more marks for his Physics and 2 more marks for his Chemistry, his average would have been 65. How many subjects were there in the examination ?
 1) 8 2) 10 3) 11 4) 14 5) None of these
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 Answer
##### Out of 300 students in a school, 95 play cricket only, 120 play football only, 80 play volleyball only and 5 play no games. If one student is chosen at random, find the probability that he plays either cricket or volleyball-
 1) 5/12 2) 7/12 3) 11/12 4) 3/11 5) None of these
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 Answer
##### The 3rd and 8th term of an arithmetic progression are -13 and 2 respectively. What is the 14th term?
 1) 18 2) 20 3) 28 4) 32 5) None of these
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 Answer
##### What is the sum of the first 12 terms of an arithmetic progression if the first term is -19 and last term is 36 ?
 1) 84 2) 102 3) 106 4) 108 5) None of these
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
##### What is the sum of the first 17 terms of an arithmetic progression if the first term is -20 and last term is 28 ?
 1) 68 2) 70 3) 72 4) 76 5) None of these
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
##### If the 3rd and the 5th term of an arithmetic progression are 13 and 21, what is the 13th term ?
 1) 53 2) 57 3) 63 4) 72 5) None of these
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 Answer
##### In an AP, the ratio of the 7th term to the 10th term is -1. If the 16th term is -15, what is the 3rd term ?
 1) 11 2) 13 3) -11 4) -13 5) None of these
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
##### How many terms of the A.P.  6,  12,  18,  24, …… must be taken to make the sum 816 ?
 1) 14 2) 16 3) 17 4) 20 5) None of these
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 Answer
##### If 7 times the seventh term of an Arithmetic Progression (AP) is equal to 11 times its eleventh term, then the 18th term of the AP will be -
 1) 1 2) 0 3) 12 4) 18 5) None of these
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 Answer
##### The 7th and 12th term of an arithmetic progression are -15 and 5 respectively. What is the 16th term ?
 1) 18 2) 19 3) 21 4) 24 5) None of these View Answer