SBI PO 2016 PYQs

Question 1

A number is such that when it is multiplied by 6, it gives another number which is more than 168 as the original number itself is less than 168. What is 15% of the original number?

Accepted Answer

Let the number be x. The problem states 6x - 168 = 168 - x. Solving this gives 7x = 336, so x = 48. Then, 15% of 48 is (15/100) * 48 = 7.2.

Question 2

I.10x2– x – 24 = 0II.y2– 2y = 0

Accepted Answer

Solve both quadratic equations. For I, factorize 10x² – x – 24 = 0 to get x = 8/5, -3/2. For II, factorize y² – 2y = 0 to get y = 0, 2. Compare all possible pairs of x and y values to find the relationship.

Question 3

9 16 25 38 57 ?

Accepted Answer

The differences between consecutive terms are 7, 9, 13, 19. The differences of these differences are 2, 4, 6. The next difference should be 8, making the next term 19+8=27. So, 57+27=84.

Question 4

?% of (140.06 × 7.99 – 679.92) = 330.01

Accepted Answer

Approximate the values: (140 * 8 - 680) = (1120 - 680) = 440. So, ?% of 440 = 330. Thus, (330/440) * 100 = 75%.

Question 5

Dharma invested Rs. P for 3 years in scheme A which offered 12% per annum simple interest. She also invested Rs. (P + 400) in scheme B which offered 10% compound interest (compounded annually), for 2 years. If the amount received from scheme A was less than that received from scheme B, by Rs. 304, what is the value of P?

Accepted Answer

Calculate Amount from Scheme A: P + (P * 12 * 3)/100 = 1.36P. Calculate Amount from Scheme B: (P+400) * (1 + 10/100)^2 = (P+400) * 1.21. Set up equation: 1.21(P+400) - 1.36P = 304. Solve for P: 1.21P + 484 - 1.36P = 304 => -0.15P = -180 => P = 1200.

Question 6

I.x2– 5x + 6 = 0II.2y2– 15y + 27 = 0

Accepted Answer

Solve x² – 5x + 6 = 0 to get x = 2, 3. Solve 2y² – 15y + 27 = 0 to get y = 3, 4.5. Comparing all possible pairs, x ≤ y is the correct relation.

Question 7

8 5 6 10 21 ?

Accepted Answer

The pattern is: (previous number * 0.5) + 1, (previous number * 1) + 1, (previous number * 1.5) + 1, (previous number * 2) + 1. So, 21 * 2.5 + 1 = 52.5 + 1 = 53.5.

Question 8

40% of 859 + 86.01 ÷ 7.99 = ?

Accepted Answer

Approximate 40% of 860 as 344. Approximate 86.01 ÷ 7.99 as 86 ÷ 8 = 10.75. Summing these gives 344 + 10.75 = 354.75, which is closest to 355.

Question 9

The number of students (both male and female) who are learning freestyle in schools C and D together is what percent less than the number of students(both male and female)who are learning the same in schools A and B together?

Accepted Answer

Calculate total freestyle students for C+D and A+B. Find the difference. Divide the difference by A+B total and multiply by 100 to get the percentage less.

Question 10

I. 3x + 2y = 301II. 7x – 5y = 74

Accepted Answer

Multiply eq (I) by 5 and eq (II) by 2. Add the new equations to eliminate y, then solve for x. (15x + 10y = 1505) + (14x - 10y = 148) => 29x = 1653 => x = 57.

Question 11

8 ? 116 152 176 188

Accepted Answer

The differences between consecutive numbers are 36, 24, 12. This is an arithmetic progression with a common difference of -12. So, the missing difference is 36 + 12 = 48. Thus, 8 + 48 = 68.

Question 12

In a bag there are 7 red balls and 5 green balls. Three balls are picked at random. What is the probability that two balls are red and one is green in colour?

Accepted Answer

Total balls = 12. Total ways to pick 3 balls = C(12,3) = 220. Ways to pick 2 red and 1 green = C(7,2) * C(5,1) = 21 * 5 = 105. Probability = 105/220 = 21/44.

Question 13

What is the average number of students learning dance forms (other than freestyle) in schools B, C and D?

Accepted Answer

Calculate total students learning non-freestyle dance in B, C, D. Sum (B: 200+220+180) + (C: 210+190+230) + (D: 190+210+200) = 1860. Divide by 9 (3 schools * 3 forms) to get 206.66, which rounds to 207.

Question 14

I.14x2– 37x + 24 = 0II.28y2– 53y + 24 = 0

Accepted Answer

Solve both quadratic equations by factorization. For I, x = 8/7, 3/2. For II, y = 8/7, 3/4. Comparing values, x can be equal to y (8/7) or greater than y (3/2 > 3/4). Thus, x ≥ y.

Question 15

25 26 30 39 55 ?

Accepted Answer

The differences between consecutive numbers are +1, +4, +9, +16. These are squares of 1, 2, 3, 4. The next difference should be 5^2 = 25. So, 55 + 25 = 80.

Question 16

Shyama invested Rs. P for 2 years in scheme A which offered 11% per annum simple interest. She also invested Rs.(600 + P) in scheme B which offered 20% compound interest(compounded annually), for 2 years. If the amount received from scheme A was less than that received from scheme B, by Rs. 1216, what is the value of P ?

Accepted Answer

Calculate amounts from both schemes: A = P(1 + 0.11*2) = 1.22P. B = (600+P)(1.2)^2 = (600+P)1.44. Set up equation: 1.44(600+P) - 1.22P = 1216. Solve for P: 864 + 1.44P - 1.22P = 1216 => 0.22P = 352 => P = 1600.

Question 17

Out of the number of students (both male and female) learning freestyle in school B,58of them are 15 years old or above. If out of the total students who are 15 years old or above, 30% are females, what is the number of female students learning freestyle who are below 15 years?

Accepted Answer

Assume total students in B are 240 (LCM of 8). Students >=15 are 5/8 * 240 = 150. Females >=15 are 30% of 150 = 45. Total students <15 are 240 - 150 = 90. Males >=15 are 150 - 45 = 105. Total females are 87. Females <15 are 87 - 45 = 42.

Question 18

The total number of members enrolled in gymnasium B from 1991 to 1994 is approximately what percent more than the total number of members enrolled in gymnasium A in 1993 and 1994 together is?

Accepted Answer

Calculate total members for B (1991-1994) and A (1993-1994). Find the difference, then divide by A's total and multiply by 100 to get the percentage increase. (B = 1100, A = 400, % increase = (700/400)*100 = 175%)

Question 19

What is the apporximate average number of female employees in H.R. department taking all companies together?

Accepted Answer

Calculate the total female employees in HR across all companies (20+25+20+25+20 = 110). Divide by 5 companies to get the average: 110/5 = 22. The closest option is 24.

Question 20

A vessel contains 180 litres of mixture of milk and water in the respective ratio of 13 : 5.Fifty–four litres of this mixture was taken out and replaced with 6 litres of water,what is the approximate percentage of water in the resultant mixture?

Accepted Answer

Initial milk:water is 13:5. After removing 54L, remaining mixture is 126L. Milk removed = 39L, Water removed = 15L. Add 6L water. Calculate new milk and water quantities, then find water percentage.