Quantitative aptitude Question for all banking Bank PO,Clerk,IBPS PO,Railway,SSC,PSUs,IAS,OAS ,C-SAT Exams

Mathmetics

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

Q1.	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

Answer : 20
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

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Q2.	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

Answer : 102
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

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Q3.	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

Answer : 68
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

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Q4.	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

Answer : 53
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

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Q5.	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

Answer : 11
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.

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Q6.	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

Answer : 16
Explanation :

Given,a₁= 6 and d =12-6 = 6

Since S_n=816

Sum of n terms of A.P. is given as, S_n=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

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Q7.	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

Answer : 0
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

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Q8.	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

Answer : 21
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

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Q9.	In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is2/3 of the number of students of 8 years of age which is 48. What is the total number of students in the school ?

1)	72	2)	80
3)	100	4)	120
5)	None of these

Answer : 100
Explanation :

Let, the Total Number of Students be x.

20% of Students are below 8 years of age.

The Number of Students above or equal 8 years of age =80% of x............(i)

Number of Students of 8 years of age = 48 .....(ii)

Number of Students above 8 years of age = 2/3 × Number of Students of 8 years of age.

Number of Students of 8 years of age.

Number of Students above age of 8 =2/3 × 48=32---(iii)

From equation (i), (ii) and (iii),

80% of x = 48+2/3 of 48

80% of x = 80

x = 100.

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Q10.	Alok borrowed some money at the rate of 6% p.a. for the first three years, 9% p.a. for the next five years and 13% p.a. for the period beyond eight years. If the total interest paid by him at the end of eleven years is Rs. 8160, how much money did he borrow ?

1)	Rs.8000	2)	Rs.10,000
3)	Rs.12,000	4)	Data Inadequate
5)	None of these

Answer : Rs.8000

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Mathmetics

Q1.

The 3rd and 8th term of an arithmetic progression are -13 and 2 respectively. What is the 14th term?

Q2.

What is the sum of the first 12 terms of an arithmetic progression if the first term is -19 and last term is 36 ?

Q3.

What is the sum of the first 17 terms of an arithmetic progression if the first term is -20 and last term is 28 ?

Q4.

If the 3rd and the 5th term of an arithmetic progression are 13 and 21, what is the 13th term ?

Q5.

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 ?

Q6.

How many terms of the A.P. 6, 12, 18, 24, …… must be taken to make the sum 816 ?

Q7.

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 -

Q8.

The 7th and 12th term of an arithmetic progression are -15 and 5 respectively. What is the 16th term ?

Q9.

In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is2/3 of the number of students of 8 years of age which is 48. What is the total number of students in the school ?

Q10.

Alok borrowed some money at the rate of 6% p.a. for the first three years, 9% p.a. for the next five years and 13% p.a. for the period beyond eight years. If the total interest paid by him at the end of eleven years is Rs. 8160, how much money did he borrow ?

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