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Aptitude - Basic Arithmetic Examples
Q 1 - Which of the following is the 16th term of A.P. 5, 8, 11, 14, 17, ...?
Answer - A
Explanation
Here a = 5, d = 8 - 5 = 3, n = 16 Using formula Tn = a + (n - 1)d T16 = 5 + (16 - 1) x 3 = 50
Q 2 - Which of the following term of A.P. 4, 9, 14, 19, 24, ... is 109?
Answer - C
Explanation
Here a = 4, d = 9 - 4 = 5 Using formula Tn = a + (n - 1)d Tn = 4 + (n - 1) x 5 = 109 where 109 is the nth term. => 4 + 5n - 5 = 109 => 5n = 109 + 1 => n = 110 / 5 = 22
Answer - D
Explanation
Here a = 7, d = 13 - 7 = 6, Tn = 205 Using formula Tn = a + (n - 1)d Tn = 7 + (n - 1) x 6 = 205 where 205 is the nth term. => 7 + 6n - 6 = 205 => 6n = 205 - 1 => n = 204 / 6 = 34
Q 4 - Which of the following is the first term of A.P. if 6th term is 12 and 8th term is 22?
Answer - A
Explanation
Using formula Tn = a + (n - 1)d T6 = a + (6 - 1)d = 12 ...(i) T8 = a + (8 - 1)d = 22 ...(ii) Substract (i) from (ii) => 2d = 10 => d = 5 Using (i) a = 12 - 5d = 12 - 25 = -13
Q 5 - Which of the following is the common difference of A.P. if 6th term is 12 and 8th term is 22?
Answer - B
Explanation
Using formula Tn = a + (n - 1)d T6 = a + (6 - 1)d = 12 ...(i) T8 = a + (8 - 1)d = 22 ...(ii) Substract (i) from (ii) => 2d = 10 => d = 5
Q 6 - Which of the following is the 16th term of A.P. if 6th term is 12 and 8th term is 22?
Answer - C
Explanation
Using formula Tn = a + (n - 1)d T6 = a + (6 - 1)d = 12 ...(i) T8 = a + (8 - 1)d = 22 ...(ii) Substract (i) from (ii) => 2d = 10 => d = 5 Using (i) a = 12 - 5d = 12 - 25 = -13 ∴ T16 = -13 + (16 - 1) x 5 = 75 - 13 = 62
Q 7 - Which of the following is the sum of first 17 term of A.P. 5, 9, 13, 17, ...?
Answer - D
Explanation
Here a = 5, d = 9 - 5 = 4, n = 17 Using formula Sn = (n/2)[2a + (n - 1)d] S17 = (17/2)[2 x 5 + (17 - 1) x 4] = (17/2)(10 + 64) = 17 x 74 / 2 = 629
Q 8 - Which of the following is the sum of the series 2, 5, 8, ..., 182?
Answer - A
Explanation
Here a = 2, d = 5 - 2 = 3, Tn = 182 Using formula Tn = a + (n - 1)d a + (n - 1)d = 182 => 2 + (n - 1) x 3 = 182 => 3n = 183 => n = 61. Using formula Sn = (n/2)[2a + (n - 1)d] S61 = (61/2)[2 x 2 + (61 - 1) x 3] = (61/2)(4 + 180) = 61 x 184 / 2 = 5612
Q 9 - What are the three numbers in A.P. if their sum is 15 and product is 80?
Answer - B
Explanation
Let've numbers are a - d, a and a + d Then a - d + a + a + d = 15 => 3a = 15 => a = 5 Now (a - d)a(a + d) = 80 => (5 - d) x 5 x (5 + d) = 80 => 25 - d2 = 16 => d2 = 9 => d = +3 or -3 ∴ numbers are either 2, 5, 8 or 8, 5, 2.
Q 10 - Which of the following is the 9th term of G.P. 3, 6 , 12, 18...?
Answer - B
Explanation
Here a = 3, r = 6 / 3 = 2, T9 = ? Using formula Tn = ar(n - 1) T9 = 3 x 2(9 - 1) =3 x 28 =3 x 256 =768
Q 11 - Which of the following is the first term of G.P. if 4th term is 54 and 9th term is 13122?
Answer - A
Explanation
Using formula Tn = ar(n - 1) T4 = ar(4 - 1) = 54 => ar3 = 54 ...(i) T9 = ar(9 - 1) = 13122 => ar8 = 13122 ...(ii) Dividing (ii) by (i) => r5 = 13122 / 54 = 243 = (3)5 => r = 3 Using (i) a x 27 = 54 => a = 2
Q 12 - Which of the following is the common ratio of G.P. if 4th term is 54 and 9th term is 13122?
Answer - B
Explanation
Using formula Tn = ar(n - 1) T4 = ar(4 - 1) = 54 => ar3 = 54 ...(i) T9 = ar(9 - 1) = 13122 => ar8 = 13122 ...(ii) Dividing (ii) by (i) => r5 = 13122 / 54 = 243 = (3)5 => r = 3
Q 13 - Which of the following is the 6th term of G.P. if 4th term is 54 and 9th term is 13122?
Answer - C
Explanation
Using formula Tn = ar(n - 1) T4 = ar(4 - 1) = 54 => ar3 = 54 ...(i) T9 = ar(9 - 1) = 13122 => ar8 = 13122 ...(ii) Dividing (ii) by (i) => r5 = 13122 / 54 = 243 = (3)5 => r = 3 Using (i) a x 27 = 54 => a = 2 ∴ T6 = ar(6 - 1) = 2 x (3)5 = 2 x 243 = 486
Q 14 - Sum of two numbers is 80. If three times of first number is same as five times of the second number, what are the numbers?
Answer - A
Explanation
Let the numbers are y and 80 - y. Then 3y = 5(80-y) => 8y = 400 ∴ y = 50 and second number = 80 - 50 = 30.
Q 15 - What is the number if its third is greater than its fifth by 16?
Answer - B
Explanation
Let the number be y. Then (y / 3) - (y / 5) = 16 => 5y - 3y = 16 x 15 = 240 => 2y = 240 ∴ y = 120
Q 16 - What is the largest number among the three consecutive multiples of 3 if there sum is 90?
Answer - C
Explanation
Let the numbers be 3y , 3y + 3, 3y + 6 Now 3y + 3y + 3 + 3y + 6 = 90 => 9y = 81 => y = 9 => largest number = 3y + 6 = 3 x 9 + 6 = 33
Q 17 - Find is the positive integer if fifteen times of it is less than its square by 16.
Answer - D
Explanation
Let the positive integer by y. Then y2 - 15y = 16 => y2 - 15y - 16 = 0 => y2 - 16y + y - 16 = 0 => y(y-16) + (y-16) = 0 => (y+1)(y-16)= 0 ∴ y = 16. as -1 is not a positive integer.
Q 18 - Find is the positive integer if twenty-three times of it is more than its square by 63.
Answer - A
Explanation
Let the positive integer by y. Then 23y - 2y2 = 63 => 23y - 2y2 - 63 = 0 => 2y2 - 23y + 63 = 0 => 2y2 - 14y - 9y + 63 = 0 => 2y(y-7) - 9(y-7)= 0 => (2y-9)(y-7)= 0 ∴ y = 7. as 9/2 is not an integer.
Q 19 - Find the smallest of three numbers if numbers are in ratio of 3:2:5 and sum of their squares is 1862.
Answer - B
Explanation
Let've number as 3y, 2y and 5y. Then 9y2 + 4y2 + 25y2 = 1862. => 38y2 = 1862 => y2 = 1862 / 38 = 49 => y = 7 ∴ smallest number = 2y = 2 x 7 = 14.
Q 20 - Sum of digits of a two digit number is 10. If digits are interchanged, obtained number is 54 less than original number. What is the number?
Answer - C
Explanation
Let the ten's digit is x and unit digit of number is y. Then x + y = 10 ...(i) (10x + y) - (10y - x) = 54 => 9x - 9y = 54 => x - y = 6 ...(ii) Adding (i) and (ii) 2x = 16 => x = 8 Using (i) y = 10 - x = 2 ∴ number is 82.