SBI PO 2017 PYQs

Question 1

A bag contains 25 cards (numbered 1, 2, 3, 4, ______, 25).Two cards are picked at random from the bag (one after another and without replacement).What is the probability that the first card drawn has a prime number and the second card drawn has a number which is multiple of 12?

Accepted Answer

There are 9 prime numbers (2,3,5,7,11,13,17,19,23) out of 25. The probability of the first card being prime is 9/25. Only 12 and 24 are multiples of 12. If the first card is prime, there are 24 cards left. The probability of the second card being a multiple of 12 is 2/24. Multiply these probabilities: (9/25) * (2/24) = 18/600 = 3/100.

Question 2

1884.88 ÷ 144.921 + 6.99 +(?)2= 69.09

Accepted Answer

Approximate the values: 1885 ÷ 145 + 7 + (?)² = 69. This simplifies to 13 + 7 + (?)² = 69, so 20 + (?)² = 69. Thus, (?)² = 49, and (?) = 7.

Question 3

I.x2= 121II.y2– 23y + 132 = 0

Accepted Answer

From I, x = ±11. From II, factorize y² - 23y + 132 = 0 as (y-11)(y-12)=0, so y = 11 or 12. Comparing values, x can be equal to y (when x=11, y=11) or x < y. Thus, x ≤ y.

Question 4

The sum of radius and height of right circular cylinder is 29 metre. If the total surface area of the cylinder is 1276 square metre, what is the volume of the cylinder? (in cubic metre)

Accepted Answer

Given r+h=29 and 2πr(r+h)=1276. Substitute r+h to find r=7m. Then h=22m. Volume = πr²h = (22/7) * 7² * 22 = 3388 m³.

Question 5

41% of 601 – 250.17 = ? – 77% of 910

Accepted Answer

Approximate 41% of 600 as 246 and 77% of 910 as 700.7. The equation becomes 246 - 250.17 = ? - 700.7. So, -4.17 = ? - 700.7, leading to ? ≈ 696.53, which is 690.

Question 6

A jar contains mixture of milkand water in the respective ratioof 3 : 1.125th of the mixture is taken out and 24 litre water was added to it. If the resultant ratio between milk and water in the jar was 2 : 1, what was the initial quantity of mixture in the jar? (in litre)

Accepted Answer

Let initial mixture be 100x. Milk = 75x, Water = 25x. After removing 1/25th, milk = 72x, water = 24x. Adding 24L water gives 72x : (24x + 24) = 2 : 1. Solving yields x = 2. Initial mixture = 100x = 200L.

Question 7

I.x2– 3x – 40 = 0II.y2– 15y + 54 = 0

Accepted Answer

Solve both quadratic equations for x and y. For x, roots are 8 and -5. For y, roots are 9 and 6. Compare all possible pairs to find the relationship.

Question 8

An article is marked up 40% higher than cost price but it was sold on x% discount. The shopkeeper thus gains 12%. What would be the S.P. of the article with C.P. of Rs. 120 and sold on x% profit?

Accepted Answer

Let CP be 100. MP = 140. SP = 112 (12% profit). Discount % (x) = ((140-112)/140)*100 = 20%. Now, for CP 120 and 20% profit, SP = 120 * 1.20 = 144.

Question 9

17 9 10 16.5 ? 90

Accepted Answer

The pattern is: previous term * 0.5 + 0.5, then * 1 + 1, then * 1.5 + 1.5, and so on. So, 16.5 * 2 + 2 = 33 + 2 = 35.

Question 10

I.2x2+ 7x + 5 = 0II.5y2+ 9y + 4 = 0

Accepted Answer

Solve both quadratic equations for x and y. For I, x = -1, -5/2. For II, y = -1, -4/5. Compare the values to find the relationship.

Question 11

There are 27 cards having numbers from 1 to 27. Two cards are picked at random one by one without replacement.What is the probability that sum of numbers on these two cards is odd ?

Accepted Answer

For the sum to be odd, one card must be odd and the other even. There are 14 odd and 13 even numbers. The probability is (14/27 * 13/26) + (13/27 * 14/26) = 14/27.

Question 12

7 6 10 27 ? 515

Accepted Answer

The pattern is (previous number * N) - (N-1) where N increases by 1 each time. So, 7*1-1=6, 6*2-2=10, 10*3-3=27. Thus, 27*4-4=104.

Question 13

I.x2– 21x + 38 = 0II.y2– 11y + 28 = 0

Accepted Answer

Solve both quadratic equations for their roots. For I, x = 2, 19. For II, y = 4, 7. Compare all combinations of x and y values to determine the relationship.

Question 14

33 40 29 42 25 ?

Accepted Answer

The series alternates between two patterns: -4 (33 to 29, 29 to 25) and +2 (40 to 42). Following the +2 pattern, the next number after 42 is 44.

Question 15

I.(x+13)2= 0II.y2+ y – 56 = 0

Accepted Answer

Solve (x+13)^2=0 to get x=-13. Factorize y^2+y-56=0 as (y+8)(y-7)=0, so y=-8 or y=7. Comparing x with both y values, x is less than y.

Question 16

A sum of Rs. 91,00 is borrowed at 20% per annum compounded annually. If the amount is to be paid in two years, what will be the total amount?

Accepted Answer

Calculate the amount using A = P(1 + R/100)^T. Substitute P=9100, R=20, T=2. This gives A = 9100 * (1.2)^2 = 9100 * 1.44 = 13104.

Question 17

316 307 282 233 152 ?

Accepted Answer

Find the differences between consecutive numbers: -9, -25, -49, -81. These are squares of odd numbers (3², 5², 7², 9²). The next difference will be -11² = -121. So, 152 - 121 = 31.

Question 18

A man spends 28% of his salaryon food. From the remaininghe spent16th on rent andsent38th to his mother. If heis left with Rs. 5280, what amount does he send to his mother?

Accepted Answer

Let total salary be X. After food, remaining is 0.72X. Rent is (1/6)*0.72X = 0.12X. Mother gets (3/8)*0.72X = 0.27X. Remaining is 0.72X - 0.12X - 0.27X = 0.33X. So, 0.33X = 5280, X = 16000. Mother gets 0.27 * 16000 = 4320.

Question 19

5 9 33 72 121 ?

Accepted Answer

The differences between consecutive terms are 4, 24, 39, 49. The second differences are 20, 15, 10. The next second difference should be 5. So, the next first difference is 49+5=54. The next term is 121+54=175.

Question 20

A and B together can finish a piece of work in 20 days. B and C can together finish the same piece of work in 30 days If A is twice as efficient as C, in how many days can C alone finish the same piece of work?

Accepted Answer

Let total work be 60 units. A+B do 3 units/day, B+C do 2 units/day. A=2C. Substitute A=2C into equations to find C's efficiency. C does 1 unit/day, so 60 days for 60 units.