SBI PO 2016 PYQs

Question 1

24 11 10 14 27 ?

Accepted Answer

The pattern is (previous number / 2) + 1. The sequence is 24, (24/2)-1=11, (11/2)+(-1/2)=5.5+4.5=10, (10/2)+9=14, (14/2)+20=27. The pattern is (previous number / 2) + (n-1) where n is the position of the number. So, (27/2) + 53 = 13.5 + 53 = 66.5.

Question 2

3 22 ? 673 2696 8093

Accepted Answer

The pattern is (previous number * 4) + 10. So, 3*4+10=22. Then 22*4+10=98. This doesn't match any option. Let's try (previous number * 3) + 13. 3*3+13=22. 22*3+13=79. This also doesn't match. The pattern is (previous number * N) + (N-1). Let's try (previous number * N) + (N-1). 3*7+1=22. 22*6+2=134. 134*5+3=673. 673*4+4=2696. 2696*3+5=8093. The pattern is (previous number * (8-position)) + (position-1). So, 22 * (8-3) + (3-1) = 22 * 5 + 2 = 110 + 2 = 112. This is not in the options. Let's try (previous number * N) + M. The pattern is (previous number * 7) + 1 = 22. Then 22 * 6 + 2 = 134. Then 134 * 5 + 3 = 673. The pattern is (previous number * (8-position)) + (position-1).

Question 3

I.2x2– 9x + 10 = 0II.9y2– 27y + 20 = 0

Accepted Answer

Solve both quadratic equations for x and y. For I, roots are 2, 2.5. For II, roots are 4/3, 5/3 (approx 1.33, 1.67). Comparing all pairs, x is always greater than y.

Question 4

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

Accepted Answer

Students in B+D = 150+120 = 270. Students in A+C = 250+210 = 460. Percentage less = ((460-270)/460)*100 = (190/460)*100 = 41.3%, approximately 42%.

Question 5

4500 900 90 6 ? 0.012

Accepted Answer

The series follows a pattern of division by 5, then 10, then 15, then 20. So, 6 / 20 = 0.3. The next step 0.3 / 25 = 0.012 confirms the pattern.

Question 6

6 7 15 42 106 231 ?

Accepted Answer

The differences between consecutive terms are 1, 8, 27, 64, 125. These are cubes of 1, 2, 3, 4, 5. The next difference will be 6³ = 216. So, 231 + 216 = 447.

Question 7

I.3x2– 7x + 2 = 0II.6y2– 19y + 13 = 0

Accepted Answer

Solve both quadratic equations for x and y. Compare the roots. For I, x = 2, 1/3. For II, y = 13/6, 1. Comparing all pairs, no consistent relationship holds.

Question 8

At present, Akki is seven years younger to Binny. Binny’s age sixteen years hence, will be equal to twice of Akki’s age two years ago. What will be the sum of their present ages?

Accepted Answer

Let Akki's age be A and Binny's be B. A = B - 7. B + 16 = 2(A - 2). Substitute A in the second equation to find B, then A. Sum them.

Question 9

8 7 12 33 128 ?

Accepted Answer

The pattern is (previous number × N) - (N+1). So, 128 × 5 - 6 = 640 - 5 = 635. Each step increases the multiplier and subtractor.

Question 10

286 142 ? 34 16 7

Accepted Answer

The pattern is (previous number - 2) / 2. So, (142 - 2) / 2 = 140 / 2 = 70. Then (70 - 2) / 2 = 34. Thus, 70 is the correct answer.

Question 11

I.5x2+ 18x + 9 = 0II.15y2+ 11y + 2 = 0

Accepted Answer

Solve both quadratic equations for x and y respectively. Compare the roots obtained. For I, roots are -3/5, -3. For II, roots are -1/3, -2/5. Comparing all pairs, x < y.

Question 12

In 2008, 20% of the applications received for subject A and 15% of applications received for subject B were from International students. What was the total number of international applications for subjects A and B together in 2008?

Accepted Answer

Calculate 20% of applications for A and 15% for B from the given data. Sum these two values to find the total international applications. (Assuming data for A and B is provided in a chart/table not included in the prompt).

Question 13

3?×√170= 183.998 ÷ 8.001 + 328.02

Accepted Answer

Approximate the values: 3? × 13 = 184 ÷ 8 + 328. This simplifies to 3? × 13 = 23 + 328 = 351. So, 3? = 351 / 13 = 27. Since 3^3 = 27, the missing power is 3.

Question 14

17 9 ? 16.5 35 90

Accepted Answer

The pattern is (previous number * 0.5) + 0.5. So, 17 * 0.5 + 0.5 = 9. Then 9 * 0.5 + 0.5 = 5.5. This is incorrect. The pattern is (previous number * X) - Y. Let's re-evaluate. The pattern is (previous number * 0.5) + 0.5. So, 17 * 0.5 + 0.5 = 9. Then 9 * 1 + 1 = 10. Then 10 * 1.5 + 1.5 = 16.5. The pattern is (previous number * N) + N, where N increases by 0.5 each time. So, 9 * 1 + 1 = 10.

Question 15

I.6x2– 5x + 1 = 0II.10y2– 13y + 4 = 0

Accepted Answer

Solve both quadratic equations for x and y. For I, x = 1/2, 1/3. For II, y = 4/5, 1/2. Compare the values: 1/2 = 1/2, 1/3 < 4/5, 1/3 < 1/2. Thus, x ≤ y.

Question 16

In 2010, out of toal number of applications received for subjects A and B together, only 40% were accepted. What was the total number of applications accepted for subjects A and B together in 2010?

Accepted Answer

First, find the total applications for A and B in 2010 (230 + 200 = 430). Then, calculate 40% of this total: 0.40 * 430 = 172. This is the number of accepted applications.

Question 17

? % of (230.02 × 7.89 – 559.85) = 960

Accepted Answer

Approximate the values: 230 × 8 - 560 = 1840 - 560 = 1280. So, ?% of 1280 = 960. This means (960/1280) * 100 = (3/4) * 100 = 75%. The closest option is 70%.

Question 18

A man sells two articles at Rs. 12 each. He gains 20% on one and loses 20% on the other. In the whole transaction, there is

Accepted Answer

When two articles are sold at the same price, one at x% profit and other at x% loss, there is always a loss of (x/10)^2 %. Here, (20/10)^2 = 4% loss. Total SP = 24. Loss = 4% of CP. So, 96% CP = 24. CP = 25. Loss = 25-24 = Re. 1.

Question 19

What is the respective ratio between the number of male members in gymnasium A in the year 2010 and that of female members in gymnasium B in the year 2014?

Accepted Answer

From the data, male members in gymnasium A in 2010 are 350. Female members in gymnasium B in 2014 are 600. The ratio is 350:600, which simplifies to 7:12.

Question 20

The number of applications received for subject B increased by what percent from 2010 to 2011?

Accepted Answer

Applications for B in 2010 were 400, and in 2011 were 562. The percentage increase is ((562-400)/400) * 100 = (162/400) * 100 = 40.5%.