Question 1

The Fourth term of an Arithmetic Progression is 37 and the Sixth term is 12 more than the Fourth term. What is the sum of the Second and Eight terms?

Accepted Answer

Let us assume the first number is a and common difference is d. According to question, 4th term of A.P = 37 a + ( n - 1 ) x d = 37 Put the value of a , n and d, we will get, a + (4 - 1 ) x d = 37 a + 3d = 37..................(1) sixth term is 12 more than the fourth term, 6th term = 12 + 4th term a + ( n - 1 ) x d = 12 + 37 a + ( 6- 1 ) x d = 39 a + 5d = 39................(2) subtract the equation (1) from (2) a + 5d - a - 3d = 39 - 37 5d - 3d= 2 2d = 2 d = 1 Put the value of d in equation (1), we will get a + 3 x 1 = 37 a = 37 - 3 a = 34 Second term = a + (n - 1) x d = 34 + (2 - 1) x 1 = 34 + 1 = 35 Six term = a + (n - 1) x d = 34 + (6 - 1) x 1 = 34 + 5 = 39 Sum of Second and Six term = 35 + 39 = 74 Sum of Second and Six term = 74 Answer is 74.

Question 2

How many 3-digits numbers are completely divisible by 6?

Accepted Answer

First 3-digit number divisible by 6 is 102 and last 3-digit number divisible by 6 is 996. Difference between two consecutive numbers divisible by 6 is 6. So 3-digit numbers divisible by 6 are 102,108,114, ......., 996. This is an Arithmetic Progression in which a = 102, d = 6 and l = 996. where a = First Number , l = Last Number and d = difference of two consecutive numbers. Let the number of terms be n. So Last term = tn Then tn = 996 Use the formula for n terms of arithmetic progression. ∴ a + ( n - 1) x d = 996 ⇒ 102 + (n - 1) x 6 = 996 ⇒ 6(n - 1) = 894 ⇒ (n - 1) = 149 ⇒ n = 150 ∴ Numbers of terms = 150

Question 3

On March 1st 2016 , sherry saved ₹ 1. Everyday starting from March 2nd 2016, he save ₹1 more than the previous day . Find the first date after March 1st 2016 at the end of which his total savings will be a perfect square.

Accepted Answer

According to the question, Every day adding 1 rs extra to previous day. Let us assume after n day, total saving will become perfect square. 1 + 2 + 3 + 4 + 5 + 6 +.......................+ tn. Apply the algebra A.P formula, sum of total rupees after n days = n(n+1)/2 Hit and trail method, Put the value of n = 2 , 3 , 4, 5 .... and so on to get the perfect square. If n = 2 n(n+1)/2 = 2 x 3 / 2 = 3 which is not perfect Square. If n = 3 n(n+1)/2 = 3 x 4 / 2 = 6 which is not perfect Square. If n = 4 n(n+1)/2 = 4 x 5 / 2 = 10 which is not perfect Square. If n = 5 n(n+1)/2 = 5 x 6 / 2 = 15 which is not perfect Square. If n = 6 n(n+1)/2 = 6 x 7 / 2 = 21 which is not perfect Square. If n = 7 n(n+1)/2 = 7 x 8 / 2 = 28 which is not perfect Square. If n = 8 n(n+1)/2 = 8 x 9 / 2 = 36 which is perfect Square. n(n+1)/2 should be a perfect square . The first value of n when this occurs would be for n = 8. thus , on the 8th of March of the required condition would come true.

Question 4

A man arranges to pay off a debt of ₹3600 in 40 annual installments which form an Arithmetic Progression (A.P). When 30 of the installments are paid, he dies leaving one-third of the debt unpaid . Find the value of the first installment.

Accepted Answer

Let us assume the first installment is a and difference between two consecutive installments is d. According to question, Sum of 40 Installments = 3600 Apply the formula Sum of first n terms in an Arithmetic Progression = Sn = n/2[ 2a + (n?1)d ] where a = the first term, d = common difference, n = number of terms. n/2[ 2a + (n?1)d ] = 3600 Put the value of a, n and d from question, 40/2[ 2a + (40 ?1)d ] = 3600 20[ 2a + 39d ] = 3600 [ 2a + 39d ] = 3600/20 = 180 2a + 39d = 180...........................(1) Again according to given question, After paying the 30 installments the unpaid amount = 1/3(total unpaid amount) = 3600 x 1/3 After paying the 30 installments the unpaid amount = 1200 So After paying the 30 installments the paid amount = 3600 - 1200 = 2400 Sum of 30 installments = 2400 30/2[ 2a + (30 ?1)d ] = 2400 [ 2a + (30 ?1)d ] = 2400 x 2/30 2a + 29d = 80 x 2 = 160 2a + 29d = 160..........................(2) Subtract the Eq. (2) from Eq. (1), we will get 2a + 39d - 2a - 29d = 180 - 160 10d = 20 d = 2 Put the value of d in Equation (1), we will get 2a + 39 x 2 = 180 2a = 180 - 78 a = 102/2 a = 51 The value of first installment = a = 51

Question 5

A number 15 is divided into 3 parts which are in Arithmetic Progression (A.P) and the sum of their squares is 83. What will be the smallest number?

Accepted Answer

Let us assume the second number is a and the difference between consecutive numbers is d. According to Arithmetic progression, First number = a - d Second number = a Third number = a + d According to question, Sum of the all three numbers = 15 a - d + a + a + d = 15 3a = 15 a = 5 Again according to given question, sum of square of the 3 numbers = 83 (a - d) 2 + a 2 + (a + d) 2 = 83 apply the algebra formula a 2 + d 2 - 2ad + a 2 + a 2 + d 2 + 2ad = 83 3a 2 + 2d 2 = 83 Put the value of a in above equation. 3 x 5 2 + 2d2 = 83 3 x 25 + 2d2 = 83 75 + 2d2 = 83 2d 2 = 83 - 75 2d 2 = 8 d 2 = 8/2 d 2 = 4 d = 2 Put the value of a and d in below equation. First number = a - d = 5 - 2 = 3 Second number = a = 5 Third number = a + d = 5 + 2 = 7 The smallest number is 3.

Question 6

A boy agrees to work at the rate of 1 rupees on the first day, 2 rupees on the second day. Four rupees on the third day and so on . how much will the boy get if he starts working on the 1st of February and finishes on the 20th of February?

Accepted Answer

According to given question, First day 1 rs, second day 2 rs, third day 4 rs..... 1 , 2 , 4............ and so on The given series is in geometric progression (GP). 1, 21, 22,............. Sum of first n terms in a geometric progression(GP) Sn = a (rn ? 1) /(r ?1) if r > 1 and where a= the first term, r = common ratio,n = number of terms As per the given question a = 1 , r = 2 and n = 20; Sn = 1(220 ? 1) /(2 ?1) Sn = (220 ? 1) /1 Sn = (220 ? 1) Sn = 220 ? 1

Question 7

what is the sum of all the two-digits numbers which when divided by 7 gives a remainder of 3?

Accepted Answer

According to question, First Two Digit Number, which is divided by 7 and give the remainder 3, will be 10. Last Two Digit Number, which is divided by 7 and give the remainder 3, will be 94. The common difference between two consecutive numbers will be 7. This series will be like →10, 17, 21 ,............................ 94. Here First Number a = 10, Last number tn = 94, Common Difference d = 7 ; Using the Formula for Last Number tn = a + (n - 1) x d; ⇒ a + (n - 1) x d = tn Put the value of a , d and tn in above equation. ⇒ 10 + (n - 1) x 7 = 94 ⇒ 10 + 7n - 7 = 94 ⇒ 7n + 3 = 94 ⇒ 7n = 94 - 3 = 91 ⇒ n = 91/7 ⇒ n = 13 Using the formula for the sum of Arithmetic Progression. Sn = n/2 [ First Number + Last Number ] ; Put the value of n , First Number and Last Number, we will get, Sn = 13/2 [ 10 + 94 ] Sn = 104 x 13/2 Sn = 52 x 13 Sn = 676

Question 8

If the m th term of an Arithmetic Progression (AP) is 1/n and n th term is 1/m, then find the sum of mn terms.

Accepted Answer

let us assume a be the first term and d the common difference. According to Question, mth term of an Arithmetic Progression (AP) is 1/n Use the Arithmetic Progression formula for mth term, ⇒ a + ( m - 1 ) x d = tm ⇒ a + ( m - 1 ) x d = 1/n ⇒ a + md - d = 1/n .......................(1) Use the Arithmetic Progression formula for nth term, ⇒ a + ( n - 1 ) x d = tn ⇒ a + ( n - 1 ) x d = 1/m ⇒ a + nd - d = 1/m.........................(2) Subtracts Equation (1) from Equation (2) , we will get ⇒ a + nd - d - (a + md - d) = 1/m - 1/n ⇒ a + nd - d - (a + md - d) = 1/m - 1/n ⇒ a + nd - d - a - md + d = 1/m - 1/n ⇒ nd - md = 1/m - 1/n ⇒ d(n - m) = 1/m - 1/n ⇒ d(n - m) = (n - m)/mn ⇒ d = 1/mn.......................................(3) Put the value of d in Equation (1), we will get, ⇒ a + md - d = 1/n ⇒ a + ( m x 1/mn ) - 1/mn = 1/n ⇒ a + 1/n - 1/mn = 1/n ⇒ a = 1/n - 1/n + 1/mn ⇒ a = 1/mn.....................................(4) Now according to question, Sum of mn terms = mn/2(2a + ( mn - 1) x d) Smn = mn/2(2a + ( mn - 1) x d) Put the value of a and d in above equation, we will get, Smn = mn/2 [ 2 x 1/mn + ( mn - 1) x 1/mn ] Smn = mn/2 [2/mn + 1 - 1/mn ] Smn = mn/2 [ 1 + 1/mn ] Smn = mn/2 [ ( mn + 1 )/mn ] Smn = mn/2 (1 + mn)/mn Smn = 1/2 (1 + mn) Smn = (1 + mn)/2 = (mn + 1)/2

Question 9

The Sum of all terms of the arithmetic progression having 10th terms except for the 1st term, is 99, and except for the 6th term , 89. Find the 3rd terms of the progression if the sum of the 1st and the 5th term is equal to 10.

Accepted Answer

Let us assume the first term is a and common difference is d. Formula for nth term of Arithmetic Progression = a + ( n - 1 ) x d So first term t1 = a + ( n - 1 ) x d t1 = a + (1 - 1) x d = a + 0 x d = a Fifth term t5 = a + ( n - 1 ) x d = a + (5 - 1) x d = a + 4d According to question, Sum of the first term and the fifth term = 10 t1 + t5 = 10 ⇒ a + a + 4d = 10 ⇒ 2a + 4d = 10 ⇒ a + 2d = 5.............................(1) Formula of sum of n terms for Arithmetic Progression, Sn = n/2[ 2a + ( n - 1 ) x d ] Put the value of n = 10, Since total number of terms is 10 . ⇒ S10 = 10/2[ 2a + ( 10 - 1 ) x d ] ⇒ S10 = 5[ 2a + 9 x d ] = 10a + 45d Again According to question, The sum of all terms of the Arithmetic Progression, except for the 1st term = 99 S10 - first term = 99 ⇒ 10a + 45d - a = 99 ⇒ 9a +45d = 99 ⇒ a + 5d = 11.................................(2) Subtracts Equation (1) from Equation (2), we will get, ⇒ a + 5d - a - 2d = 11 - 5 ⇒ 3d = 6 ⇒ d = 2 Put the value of d in equation (1), we will get ⇒ a + 2 x 2 = 5 ⇒ a + 4 = 5 ⇒ a = 5 - 4 ⇒ a = 1 Since tn = a + ( n - 1) x d find the third term by putting the value of a , n and d. we will get, t3 = 1 + ( 3 - 1) x 2 = 1 + 2 x 2 = 1 + 4 = 5 So 3rd term of Arithmetic Progression is 5.

Question 10

If 24 is subtracted from a number, it reduces to its four-seventh. what is the sum of the digits of that number?

Accepted Answer

Let us assume the number is n. According to question, If 24 is subtracted, it becomes 4/7 of the number. n - 24 = n x 4/7 ⇒ n - 4n/7 = 24 ⇒ (7n - 4n)/7 = 24 ⇒ 3n/7 = 24 ⇒ 3n = 24 x 7 ⇒ n = 24 x 7/3 ⇒ n = 8 x 7 ⇒ n = 56 Sum of the digits of the number = 5 + 6 = 11

Question 11

If the sum of one-half and one-fifth of the number exceeds one-third of that number by 7 1/3 , the number is

Accepted Answer

Let us assume the number is a. Given that 7 1/3 = 22/3 According to question, a x 1/2 + a x 1/5 = a x 1/3 + 22/3 ⇒ a/2 + a/5 = a/3 + 22/3 ⇒ a/2 + a/5 - a/3 = 22/3 ⇒ (15a + 6a - 10a)/30 = 22/3 ⇒ (15a + 6a - 10a) = 30 x 22/3 ⇒ 11a = 10 x 22 ⇒ a = 10 x 2 ⇒ a = 20

Question 12

A driver's income consists of his salary and tips. during one week his tips were 5/4 of his salary. what fraction of his income came from tips?

Accepted Answer

Let us assume the salary of driver be ₹ R Then Tips of week = R x 5/4 According to question, his income during one week = Salary + Tips ⇒ Total Income during one week = R + (5R/4) ⇒ Total Income during one week = (4R + 5R)/4 = 9R/4 Required Fraction = Tips in a week/Total Income in a week ⇒ Required Fraction = 5R/4 / 9R/4 ⇒ Required Fraction = 5R/4 x 4/9R ⇒ Required Fraction = 5/9

Question 13

In the certain party, there was a bowl of rice for every two guests , a bowl of juice for every three of them and a bowl of meat for every four of them. If there were all 65 bowls of food , then how many guests were there in the party?

Accepted Answer

Let the number of rice bowls be a, the number of juice bowls be b, and the number of meat bowls be c. According to question, a + b + c = 65........................(1) The total number of guests = 2a The total number of guests = 3b The total number of guests = 4c So the total number of guests will be same in the party. 2a = 3b = 4c..........................(2) As per Equation (2) b = 2a/3................................(3) c = 2a/4 = a/2......................(4) Now put the value of b and c from Equation (3), (4) in Equation (1), a + 2a/3 + a/2 = 65 (6a + 4a + 3a)/6 = 65 13a = 65 x 6 a = 5 x 6 = 30 Put the value of a in equation (3) and (4) in order to get the value of b and c, b = 2 x 30/3 = 2 x 10 = 20 c = 30/2 = 15 The Total number of Guests = 2a = 3b = 4c = 60

Question 14

Ram and Mohan are friends. Each has some money. If Ram gives ₹ 30 to Mohan, Then Mohan will have twice the money left with Ram. But if Mohan gives ₹ 10 to Ram, Then Ram will have thrice as much as is left with Mohan. How much money does each have?

Accepted Answer

Let us assume Ram has R rupees and Mohan has M rupees. According to question, If Ram gives 30 rupees to Mohan, then Ram has left money = R - 30 and Mohan has money = M + 30 Then Mohan will have twice the money left with Ram, M + 30 = 2(R - 30) M + 30 = 2R - 60 2R - M = 90................................(1) Again According to question, if Mohan gives 10 rupees to Ram, then Mohan has left the money = M - 10 and Ram has the money = R + 10 Then According to question, Ram will have thrice as much as is left with Mohan, R + 10 = 3 (M - 10 ) ⇒ R + 10 = 3M - 30 ⇒ 3M - R = 10 + 30 ⇒ 3M - R = 40..................................(2) After Multiplying 2 with Equation (2) , add with the equation (1), 6M - 2R + 2R - M = 80 + 90 ⇒ 6M - M = 170 ⇒ 5M = 170 ⇒ M = 170/5 ⇒ M = 34 Put the value of M in equation (1), we will get ⇒ 2R - 34 = 90 ⇒ 2R = 90 + 34 ⇒ 2R = 124 ⇒ R = 124/2 ⇒ R = 62

Question 15

The sum of two numbers is 25 and their difference is 13. Find their product.

Accepted Answer

Let us assume the numbers are a and b. According to question, sum of two numbers = 25 a + b = 25......................(1) difference of two numbers = 13 a - b = 13........................(2) add the Equation (1) and (2) a + b+ a - b = 25 + 13 ⇒ 2a = 38 ⇒ a = 19 Put the value of a in equation in (1) 19 + b = 25 ⇒ b = 25 - 19 ⇒ b = 6 Product of the numbers = a x b put the value of a and b, ⇒ Product of the numbers = 19 x 6 ⇒ Product of the numbers = 114

Question 16

The difference between the squares of two numbers is 256000 and the sum of the numbers is 1000. The numbers are

Accepted Answer

Let us assume the numbers are a and b. According to question, a2 - b2 = 256000 .......................(1) and a + b = 1000 .........................(2) On dividing the Equation (1) with Equation (2), (a2 - b2)/(a + b) = 256000/1000 (a2 - b2)/(a + b) = 256 (a + b)(a - b)/(a + b) = 256 a - b = 256...............................(3) Add the equation (2) and (3), we will get a + b + a - b = 1000 + 256 2a = 1256 a = 628 Put the value of a in equation (2), we will get 658 + b = 1000 b = 1000 - 628 b = 372 So answer is 628 , 372.

Question 17

Two-fifths of one -fourth of three -seventh of a number is 15. What is half of that number?

Accepted Answer

Let us assume the number is P. According the question, three -seventh of a number = 3/7 x P one -fourth of three -seventh of a number = 1/4 x 3/7 x P Two-fifths of one -fourth of three -seventh of a number = 2/5 x 1/4 x 3/7 x P ⇒ 2/5 x 1/4 x 3/7 x P = 15 ⇒ 2 x 1 x 3 x P = 15 x 7 x 4 x 5 ⇒ P = 15 x 7 x 4 x 5 / 2 x 3 ⇒ P = 5 x 7 x 2 x 5 ⇒ P = 350 ∴ P /2 = 350/2 ⇒ P /2 = 175

Question 18

The numbers are such that the square of one is 224 less than 8 times the square of the other. If the numbers be in the ratio 3 : 4, the numbers are

Accepted Answer

Let us assume the ratio term is r. then the number will be 3r and 4r. According to question, (4r)2 + 224 = 8 x (3r)2 16r2 + 224 = 8 x 9r2 16r2 + 224 = 72r2 72r2 - 16r2 = 224 56r2 = 224 r2 = 224/56 r2 = 4 r = 2 First number = 3r = 3 x 2 = 6 Second number = 4r = 4 x 2 = 8

Question 19

A person on tour has total ₹360 for his daily expenses. He decides to extend his tour programme by 4 days which leads to cutting down daily expenses by ₹3 a day . The numbers of days of tour are :

Accepted Answer

Method 1 to solve this question. Let us assume the numbers of days of tour before extending the tour are D and daily expenses is E. According to question, Total expenses on tour = per day expenses x total number of days Total expenses on tour = E x D = ED 360 = ED ⇒ ED = 360 ⇒ E = 360/D .................(1) Again according to question, After extending the tour by 4 days, Number of days of tour = D + 4 Expenses will be reduced by 3 rupees, then everyday expenses = E - 3 so total expenses = (D + 4) x (E - 3) 360 = (D + 4) x (E - 3) (D + 4) x (E - 3) = 360 .............................................(2) Put the value of E in Equation (2). we will get, (D + 4) x (360/D - 3) = 360 ⇒ (D + 4) x ( (360 - 3D)/D ) = 360 ⇒ (D + 4) x ( (360 - 3D) ) = 360 x D ⇒ (D + 4) x (360 - 3D) = 360 x D After multiplication by algebra law, ⇒ 360 x D - 3D x D + 4 x 360 - 4 x 3D = 360D ⇒ 360 x D - 3D + 1440 - 12 D = 360 D ⇒ - 3 D + 1440 - 12 D = 360 D - 360 x D ⇒ - 3 D + 1440 - 12 D = 0 ⇒ 3 D - 1440 + 12 D = 0 ⇒ D - 480 + 4 D = 0 ⇒ D + 4 D - 480 = 0 ⇒ D + 24 D - 20 D - 480 = 0 ⇒ D( D + 24) - 20( D + 24) = 0 ⇒ ( D + 24) ( D - 20) = 0 Either ( D + 24) = 0 or ( D - 20) = 0 So D = - 24 or D = 20 But days cannot be negative so D = 20 days. Method 2 to solve this question. Let us assume the original day of tour is d days. Given that his tour is extended for 4 days Hence daily expenses per days = 360/(d + 4) Therefore, According to question, 360/d - 360/(d + 4) = 3 ⇒ (360(d + 4) - 360d)/d x (d + 4) = 3 ⇒ (360(d + 4) - 360d)/d x (d + 4) = 3 ⇒ 360(d + 4) - 360d = 3d x (d + 4) ⇒ 360d + 1440 – 360d = 3(d 2 + 4d) ⇒ 1440 = 3d 2 + 12d ⇒ 3d 2 + 12d – 1440 = 0 ⇒ d 2 + 4d – 480 = 0 ⇒ d 2 + 24d – 20d – 480 = 0 ⇒ d(d + 24) – 20(d + 24) = 0 ⇒ (d + 24)(d – 20) = 0 ⇒ (d + 24) = 0 or (d – 20) = 0 ? d = –24 or d = 20 Since days cannot be negative, d = 20 Hence his original duration of the tour is 20 days. <> <> <> <> <> <> <>

Question 20

The sum of three consecutive odd numbers is 20 more than the first of these numbers.What is the middle number?

Accepted Answer

Let us assume the first Odd number is a. then 2nd odd number = a + 2 and 3rd odd number = a + 4 According to question, Sum of three consecutive odd numbers is 20 more than the first of these numbers; a + (a + 2) + (a + 4) = a + 20 ⇒ a + a + 2 + a + 4 - a = 20 ⇒ 2a + 6 = 20 ⇒ 2a + 6 = 20 ⇒ 2a = 20 - 6 ⇒ 2a = 14 ⇒ a = 7 The First odd number a = 7 The second odd number = a + 2 = 7 + 2 = 9 Second odd number is middle number = 9

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