Free Mock Test SBI PO 2017. Mock test for SBI PO Prelim 2017

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We are providing you with Quant Section Mock for the upcoming SBI PO 2017 Prelim Exam. It contains 35 questions and time limit is 24 minutes.

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Question 1 of 35

1. Question
1 points
Directions (1-5): What will come in place of question mark (?) in the following number series?

4, 14, 31, 95, 193, ?
- 465
- 581
- 532
- 567
- 483
Correct

*3 + 2, *2 + 3, *3 + 2, *2 + 3, *3 + 2

Incorrect

*3 + 2, *2 + 3, *3 + 2, *2 + 3, *3 + 2

Unattempted

*3 + 2, *2 + 3, *3 + 2, *2 + 3, *3 + 2
Question 2 of 35

2. Question
1 points
Directions (1-5): What will come in place of question mark (?) in the following number series?

2, 3, 4, 20, ?, 2448
- 176
- 243
- 152
- 255
- 287
Correct

*1 + 1, *2 – 2, *4 + 4, *8 – 8, *16 + 16.

Incorrect

*1 + 1, *2 – 2, *4 + 4, *8 – 8, *16 + 16.

Unattempted

*1 + 1, *2 – 2, *4 + 4, *8 – 8, *16 + 16.
Question 3 of 35

3. Question
1 points
Directions (1-5): What will come in place of question mark (?) in the following number series?

9, 5, 9, 25, 91, ?
- 388
- 365
- 414
- 454
- 408
Correct

*0.5 + 0.5, *1.5 + 1.5, *2.5 + 2.5, *3.5 + 3.5, *4.5 + 4.5.

Incorrect

*0.5 + 0.5, *1.5 + 1.5, *2.5 + 2.5, *3.5 + 3.5, *4.5 + 4.5.

Unattempted

*0.5 + 0.5, *1.5 + 1.5, *2.5 + 2.5, *3.5 + 3.5, *4.5 + 4.5.
Question 4 of 35

4. Question
1 points
Directions (1-5): What will come in place of question mark (?) in the following number series?

5, 10, 22, 48, 95, ?
- 170
- 187
- 155
- 190
- 209
Correct

Difference of difference series.

Incorrect

Difference of difference series.

Unattempted

Difference of difference series.
Question 5 of 35

5. Question
1 points
Directions (1-5): What will come in place of question mark (?) in the following number series?

3, 5, 13, 43, 177, ?
- 845
- 833
- 923
- 891
- None of these
Correct

*1 + 2, *2 + 3, *3 +4, *4 + 5, *5 + 6

Incorrect

*1 + 2, *2 + 3, *3 +4, *4 + 5, *5 + 6

Unattempted

*1 + 2, *2 + 3, *3 +4, *4 + 5, *5 + 6
Question 6 of 35

6. Question
1 points
Directions (6- 10): What approximate value should come in place of the question mark (?) in the following questions?

46.09% of 949.89 + 2.5 * 2.4 = 23.123 + ?
- 445
- 325
- 420
- 380
- 480
Correct

Incorrect

Unattempted
Question 7 of 35

7. Question
1 points
Directions (6- 10): What approximate value should come in place of the question mark (?) in the following questions?

(61.87)² – √11020 = (?)² + 375.22
- 58
- 52
- 63
- 48
- 61
Correct

Incorrect

Unattempted
Question 8 of 35

8. Question
1 points
Directions (6- 10): What approximate value should come in place of the question mark (?) in the following questions?

4626.4 + 629.06 = (?)% of 6600 + 3473.46
- 35
- 39
- 19
- 15
- 27
Correct

Incorrect

Unattempted
Question 9 of 35

9. Question
1 points
Directions (6- 10): What approximate value should come in place of the question mark (?) in the following questions?

44.92% of 640.08 – ?% of 959.95 = 48.126
- 18
- 34
- 39
- 25
- 9
Correct

Incorrect

Unattempted
Question 10 of 35

10. Question
1 points
Directions (6- 10): What approximate value should come in place of the question mark (?) in the following questions?

(10.02 × 15.2 × 17.99)^1/2 + 90.04 × 22.82 ÷ 10 – 20.08% of 144.98 = ?
- 215
- 287
- 230
- 250
- 310
Correct

Incorrect

Unattempted
Question 11 of 35

11. Question
1 points
An article is sold making a loss of 10%. Had it been bought for 20% less and sold for Rs 18 more, there would have been a profit of 25%. What is the cost price of the article?
- Rs 120
- Rs 145
- Rs 230
- Rs 180
- None of these
Correct

Let CP = Rs x
So SP = (90/100)*x = 9x/10
Now if CP = (80/100)*x = 4x/5
And SP is (9x/10 + 18), the profit is 25%
So (9x/10 + 18 – 4x/5)/(4x/5) * 100 = 25
Solve, x = Rs 180

Incorrect

Let CP = Rs x
So SP = (90/100)*x = 9x/10
Now if CP = (80/100)*x = 4x/5
And SP is (9x/10 + 18), the profit is 25%
So (9x/10 + 18 – 4x/5)/(4x/5) * 100 = 25
Solve, x = Rs 180

Unattempted

Let CP = Rs x
So SP = (90/100)*x = 9x/10
Now if CP = (80/100)*x = 4x/5
And SP is (9x/10 + 18), the profit is 25%
So (9x/10 + 18 – 4x/5)/(4x/5) * 100 = 25
Solve, x = Rs 180
Question 12 of 35

12. Question
1 points
Ages of Avni and Bhumika is in the ratio 2 : 3 while that of Avni and Charru is in the ratio 4 : 9. If the difference in the ages of Avni and Charu is 25 years, what is the sum of ages of Avni and Bhumika?
- 46
- 50
- 60
- 40
- 52
Correct

B/A = 3/2 and A/C = 4/9
So B : A : C = 3*4 : 2*4 : 2*9 = 6 : 4 : 9
So (9-4)/(6+4+9) * x = 25
Solve, x = 95
Sum of ages of A and B = (6+4)/(6+4+9) * 95 = 50

Incorrect

B/A = 3/2 and A/C = 4/9
So B : A : C = 3*4 : 2*4 : 2*9 = 6 : 4 : 9
So (9-4)/(6+4+9) * x = 25
Solve, x = 95
Sum of ages of A and B = (6+4)/(6+4+9) * 95 = 50

Unattempted

B/A = 3/2 and A/C = 4/9
So B : A : C = 3*4 : 2*4 : 2*9 = 6 : 4 : 9
So (9-4)/(6+4+9) * x = 25
Solve, x = 95
Sum of ages of A and B = (6+4)/(6+4+9) * 95 = 50
Question 13 of 35

13. Question
1 points
Raghav borrowed a sum of Rs 5000 from his father at a certain rate of interest. After 2 years, he borrowed a sum of Rs 9000 again from his father at same interest rate. If after a further of 2 years he paid a total of Rs 18,560 to his father, what is the rate of percent at which he borrowed the money?
- 11 2/3%
- 11%
- 10.5%
- 12%
- 14%
Correct

Total principal = 5000+9000 = 14000
So interest = 18560 – 14000 = 4560
So (5000*r*4)/100 + (9000*r*2)/100= 4560
Solve, r= 12%

Incorrect

Total principal = 5000+9000 = 14000
So interest = 18560 – 14000 = 4560
So (5000*r*4)/100 + (9000*r*2)/100= 4560
Solve, r= 12%

Unattempted

Total principal = 5000+9000 = 14000
So interest = 18560 – 14000 = 4560
So (5000*r*4)/100 + (9000*r*2)/100= 4560
Solve, r= 12%
Question 14 of 35

14. Question
1 points
What is the probability of drawing 2 cards from a pack of 52 cards such that there is at least 1 king in the draw?
- 46/221
- 53/221
- 23/221
- 33/221
- None of these
Correct

Case 1: 1 is king
⁴C₁*⁴⁸C₁ / ⁵²C₂
Case 2: both are king
⁴C₂ / ⁵²C₂
Add both cases.

Incorrect

Unattempted
Question 15 of 35

15. Question
1 points
From her monthly salary Sakshi spent 25% in shopping. Out of the remaining salary spent 20% on food and 4% on her cosmetics. If she is left with Rs 57,600 at the end of month, what is her monthly salary?
- Rs 2.5 lakh
- Rs 2.3 lakh
- Rs 1 lakh
- Rs 1.7 lakh
- Rs 0.8 lakh
Correct

Let x is monthly salary
So 25% in shopping, remaining 75%
20% of 75% on furniture, so left with 80% of 75%
Now spent 4% of 80% of 75% on food
So left with 96% of 80% of 75% of x which is 57600
So (96/100)*(80/100)*(75/100)*x = 57,600
Solve, x = 100,000

Incorrect

Let x is monthly salary
So 25% in shopping, remaining 75%
20% of 75% on furniture, so left with 80% of 75%
Now spent 4% of 80% of 75% on food
So left with 96% of 80% of 75% of x which is 57600
So (96/100)*(80/100)*(75/100)*x = 57,600
Solve, x = 100,000

Unattempted

Let x is monthly salary
So 25% in shopping, remaining 75%
20% of 75% on furniture, so left with 80% of 75%
Now spent 4% of 80% of 75% on food
So left with 96% of 80% of 75% of x which is 57600
So (96/100)*(80/100)*(75/100)*x = 57,600
Solve, x = 100,000
Question 16 of 35

16. Question
1 points
Directions (16-20): Study the following table carefully to answer the following questions.

The table shows the number of students in Electronics and Computer department over the given years and the ratio of males to females in both departments over the given years

If the number of students in Computer Department in 2012 is 350 and the ratio of males to females in Computer Department in 2013 is 1 : 1, then what is the number of males in Computer department over these years?
- 460
- 450
- 380
- 410
- 360
Correct

(3/5)*350 + (1/2)*500

Incorrect

(3/5)*350 + (1/2)*500

Unattempted

(3/5)*350 + (1/2)*500
Question 17 of 35

17. Question
1 points
Directions (16-20): Study the following table carefully to answer the following questions.

The table shows the number of students in Electronics and Computer department over the given years and the ratio of males to females in both departments over the given years

If the number of students in Computer and Electronics department is 500 and 450 resp. in year 2010 and the ratio of males to females in Computer department in 2013 is same as given in question 21, then number of males in Electronics department in 2010 and 2013 together is approximately what percent more than the number of females in Computer department in the same years?
- 8%
- 10%
- 5%
- 12%
- 6%
Correct

Females in C in 2010 and 2013 = (3/5)*500 + (1/2)*500 = 550
Males in E in 2010 and 2013 = (7/10)*450 + (2/5)*700 = 595
So required % = (595-550)/550 * 100

Incorrect

Females in C in 2010 and 2013 = (3/5)*500 + (1/2)*500 = 550
Males in E in 2010 and 2013 = (7/10)*450 + (2/5)*700 = 595
So required % = (595-550)/550 * 100

Unattempted

Females in C in 2010 and 2013 = (3/5)*500 + (1/2)*500 = 550
Males in E in 2010 and 2013 = (7/10)*450 + (2/5)*700 = 595
So required % = (595-550)/550 * 100
Question 18 of 35

18. Question
1 points
Directions (16-20): Study the following table carefully to answer the following questions.

The table shows the number of students in Electronics and Computer department over the given years and the ratio of males to females in both departments over the given years

In which of the following condition, the number of females is greater?
- in Computer department in 2010 (given number of students in department is 500)
- in Electronics department in 2010 (given number of students in department is 450)
- in Computer department in 2012 (given number of students in department is 350)
- in Electronics department in 2014 (given ratio of males to females in department is 8 : 5)
- in Computer department in 2013 (given ratio of males to females in department is 1 : 1)
Correct

1) (3/5)*500 = 300
2) (3/10)*450 = 135
3) (2/5)*350 = 140
4) (5/13)*650 = 250
5) (1/2)*500 = 250

Incorrect

1) (3/5)*500 = 300
2) (3/10)*450 = 135
3) (2/5)*350 = 140
4) (5/13)*650 = 250
5) (1/2)*500 = 250

Unattempted

1) (3/5)*500 = 300
2) (3/10)*450 = 135
3) (2/5)*350 = 140
4) (5/13)*650 = 250
5) (1/2)*500 = 250
Question 19 of 35

19. Question
1 points
Directions (16-20): Study the following table carefully to answer the following questions.

The table shows the number of students in Electronics and Computer department over the given years and the ratio of males to females in both departments over the given years

If the number of students in Computer Department in 2012 is 350, then the number of females in Computer department in 2012 and 2014 together is approximately what percent of total students in Computer department for the same years?
- 54%
- 51%
- 46%
- 49%
- 40%
Correct

2/5)*350 + (7/13)*650 = 490
Total = 350 + 650 = 1000
So required % = (490/1000)*100

Incorrect

2/5)*350 + (7/13)*650 = 490
Total = 350 + 650 = 1000
So required % = (490/1000)*100

Unattempted

2/5)*350 + (7/13)*650 = 490
Total = 350 + 650 = 1000
So required % = (490/1000)*100
Question 20 of 35

20. Question
1 points
Directions (16-20): Study the following table carefully to answer the following questions.

The table shows the number of students in Electronics and Computer department over the given years and the ratio of males to females in both departments over the given years

What is the total number of males in both departments in 2014 if the ratio of males to females in Electronics Department in 2014 is 8 : 5?
- 650
- 700
- 770
- 710
- None of these
Correct

(6/13)*650 + (8/13)*650 = 700

Incorrect

(6/13)*650 + (8/13)*650 = 700

Unattempted

(6/13)*650 + (8/13)*650 = 700
Question 21 of 35

21. Question
1 points
Anil, Ajay and Aarti started a business by investing Rs 6000, Rs 5000 and Rs 6500 respectively. They all added the same amount as before after 4 months from the start of business. After a further of 4 months each of them took out Rs 1000. What is the ratio of their shares in total profit after a year?
- 44 : 48 : 61
- 43 : 48 : 65
- 58 : 87 : 48
- 65 : 43 : 67
- 58 : 48 : 63
Correct

6000*4 +12,000*4 + 11,000*4 : 5000*4 +10,000*4 + 9,000*4 : 6500*4 +13,000*4 + 12,000*4
60 + 120 + 110 : 50 + 100 + 90 : 65 + 130 + 120
290 : 240 : 315
58 : 48 : 63

Incorrect

6000*4 +12,000*4 + 11,000*4 : 5000*4 +10,000*4 + 9,000*4 : 6500*4 +13,000*4 + 12,000*4
60 + 120 + 110 : 50 + 100 + 90 : 65 + 130 + 120
290 : 240 : 315
58 : 48 : 63

Unattempted

6000*4 +12,000*4 + 11,000*4 : 5000*4 +10,000*4 + 9,000*4 : 6500*4 +13,000*4 + 12,000*4
60 + 120 + 110 : 50 + 100 + 90 : 65 + 130 + 120
290 : 240 : 315
58 : 48 : 63
Question 22 of 35

22. Question
1 points
Two persons P and Q start from city A at a speed of 30 km/hr and 20 km/hr respectively. P reached the city B and returns back and meets Q at point S. If the cities A and B are 90 km apart, find the distance between points A and S.
- 56 km
- 60 km
- 50 km
- 72 km
- 84 km
Correct

Time taken by P to reach city B is 3 hr. In 3 hr, distance covered by Q is 60 km. Now after this time taken by both to reach point S will be equal. Let from person B at x distance they meet, then A covered (90-60)-x = (30-x). So
x/20 = (30-x)/30.
Solve, x = 12
So distance A to S is 60 + 12

Incorrect

Time taken by P to reach city B is 3 hr. In 3 hr, distance covered by Q is 60 km. Now after this time taken by both to reach point S will be equal. Let from person B at x distance they meet, then A covered (90-60)-x = (30-x). So
x/20 = (30-x)/30.
Solve, x = 12
So distance A to S is 60 + 12

Unattempted

Time taken by P to reach city B is 3 hr. In 3 hr, distance covered by Q is 60 km. Now after this time taken by both to reach point S will be equal. Let from person B at x distance they meet, then A covered (90-60)-x = (30-x). So
x/20 = (30-x)/30.
Solve, x = 12
So distance A to S is 60 + 12
Question 23 of 35

23. Question
1 points
A work which can be completed by 20 men complete a work in 8 days can also be completed by 30 women in 7 days. At the start of work, there are 10 men. After they completed 3/4th of the work, 14 women also join them. In what time (approximate days) will the work get completed?
- 10 days
- 14 days
- 12 days
- 17 days
- 20 days
Correct

20 men in 8 days, so10 men in (20*8)/10 = 16 days
They completed 3/4 of the work so days required by them is (3/4) * 16 = 12 days
Now joined by 14 women, and they now have to complete remaining work i.e. 1 – (3/4) = 1/4
Let x days are required to do this 1/4 work.
30 women in 7 days, so 14 women in (30*7)/14 = 15 days
Now 10 men and 14 women have to complete 1/4 work in x days. So
[1/16 + 1/15] * x = 1/4
Solve, x= 60/31 ≈ 2
So total days 12 + 2 = 14 days

Incorrect

20 men in 8 days, so10 men in (20*8)/10 = 16 days
They completed 3/4 of the work so days required by them is (3/4) * 16 = 12 days
Now joined by 14 women, and they now have to complete remaining work i.e. 1 – (3/4) = 1/4
Let x days are required to do this 1/4 work.
30 women in 7 days, so 14 women in (30*7)/14 = 15 days
Now 10 men and 14 women have to complete 1/4 work in x days. So
[1/16 + 1/15] * x = 1/4
Solve, x= 60/31 ≈ 2
So total days 12 + 2 = 14 days

Unattempted

20 men in 8 days, so10 men in (20*8)/10 = 16 days
They completed 3/4 of the work so days required by them is (3/4) * 16 = 12 days
Now joined by 14 women, and they now have to complete remaining work i.e. 1 – (3/4) = 1/4
Let x days are required to do this 1/4 work.
30 women in 7 days, so 14 women in (30*7)/14 = 15 days
Now 10 men and 14 women have to complete 1/4 work in x days. So
[1/16 + 1/15] * x = 1/4
Solve, x= 60/31 ≈ 2
So total days 12 + 2 = 14 days
Question 24 of 35

24. Question
1 points
Circumference of a circle is 66m more than the perimeter of the rectangle. If the length and breadth of the rectangle are in the ratio 6 : 5 and the radius of circle and breadth of rectangle are in ratio 7 : 5, find the area of the rectangle.
- 365 m^2
- 230 m^2
- 195 m^2
- 270 m^2
- 225 m^2
Correct

l/b = 6/5 and b/r = 5/7
so l : b : r = 6 : 5 : 7
so 2* (22/7) * 7x = 66 + 2[6x+5x]
solve, x = 3
so area of rect = 6x * 5x = 30*9

Incorrect

l/b = 6/5 and b/r = 5/7
so l : b : r = 6 : 5 : 7
so 2* (22/7) * 7x = 66 + 2[6x+5x]
solve, x = 3
so area of rect = 6x * 5x = 30*9

Unattempted

l/b = 6/5 and b/r = 5/7
so l : b : r = 6 : 5 : 7
so 2* (22/7) * 7x = 66 + 2[6x+5x]
solve, x = 3
so area of rect = 6x * 5x = 30*9
Question 25 of 35

25. Question
1 points
A can work 2/3 times as fast as B and C together. A and C together can work twice as fast as B. If C alone takes 45 days to complete the work, how long will A take to complete the same work?
- 20 days
- 30 days
- 36 days
- 27 days
- 32 days
Correct

Days ratio –
A : (B+C) = 3 : 2 => 3x and 2x
and B : (A+C) = 2 : 1 => 2y and y
So
1/2x – 1/45 = 1/2y
and
1/y – 1/45 = 1/3x
Solve both equations, x = 10
So A will take 3x = 30 days

Incorrect

Days ratio –
A : (B+C) = 3 : 2 => 3x and 2x
and B : (A+C) = 2 : 1 => 2y and y
So
1/2x – 1/45 = 1/2y
and
1/y – 1/45 = 1/3x
Solve both equations, x = 10
So A will take 3x = 30 days

Unattempted

Days ratio –
A : (B+C) = 3 : 2 => 3x and 2x
and B : (A+C) = 2 : 1 => 2y and y
So
1/2x – 1/45 = 1/2y
and
1/y – 1/45 = 1/3x
Solve both equations, x = 10
So A will take 3x = 30 days
Question 26 of 35

26. Question
1 points
Directions (26-30): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

I. 3x² + 10x – 8 = 0,
II. 2y² – 13y + 6 = 0
- x > y
- x ≤ y
- x < y
- x ≥ y
- x = y or relation cannot be established
Correct

3x² + 10x – 8 = 0
3x² + 12x – 2x – 8 = 0
Gives x = -2, 2/3
2y² – 13y + 6 = 0
2y² – 12y – y + 6 = 0
Gives y = 1/2, 6

Incorrect

3x² + 10x – 8 = 0
3x² + 12x – 2x – 8 = 0
Gives x = -2, 2/3
2y² – 13y + 6 = 0
2y² – 12y – y + 6 = 0
Gives y = 1/2, 6

Unattempted

3x² + 10x – 8 = 0
3x² + 12x – 2x – 8 = 0
Gives x = -2, 2/3
2y² – 13y + 6 = 0
2y² – 12y – y + 6 = 0
Gives y = 1/2, 6
Question 27 of 35

27. Question
1 points
Directions (26-30): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

I. 3x² + 22 x + 24 = 0
II. 2y² + 11y + 12 = 0
- x > y
- x < y
- x = y or relation cannot be established
- x ≥ y
- x ≤ y
Correct

3x² + 22 x + 24 = 0
3x² + 18x + 4x + 24 = 0
Gives x = -4/3, -6
2y² + 11y + 12 = 0
2y² + 8y + 3y + 12 = 0
Gives y = -4, -3/2

Incorrect

3x² + 22 x + 24 = 0
3x² + 18x + 4x + 24 = 0
Gives x = -4/3, -6
2y² + 11y + 12 = 0
2y² + 8y + 3y + 12 = 0
Gives y = -4, -3/2

Unattempted

3x² + 22 x + 24 = 0
3x² + 18x + 4x + 24 = 0
Gives x = -4/3, -6
2y² + 11y + 12 = 0
2y² + 8y + 3y + 12 = 0
Gives y = -4, -3/2
Question 28 of 35

28. Question
1 points
Directions (26-30): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

I. 4x² + 3x – 27 = 0
II. 15y² – 38y – 21 = 0
- x ≤ y
- x < y
- x ≥ y
- x > y
- x = y or relation cannot be established
Correct

4x² + 3x – 27 = 0
4x² + 12x – 9x – 27 = 0
So x =2.25, -3
15 y² – 38y – 21 = 0
15 y² – 45y + 7y – 21 = 0
So y = 3, – 0.46

Incorrect

4x² + 3x – 27 = 0
4x² + 12x – 9x – 27 = 0
So x =2.25, -3
15 y² – 38y – 21 = 0
15 y² – 45y + 7y – 21 = 0
So y = 3, – 0.46

Unattempted

4x² + 3x – 27 = 0
4x² + 12x – 9x – 27 = 0
So x =2.25, -3
15 y² – 38y – 21 = 0
15 y² – 45y + 7y – 21 = 0
So y = 3, – 0.46
Question 29 of 35

29. Question
1 points
Directions (26-30): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

I. 4x² + 19x + 21 = 0
II. 2y² – 25y – 27 = 0
- x > y
- x = y or relation cannot be established
- x < y
- x ≥ y
- x ≤ y
Correct

4x² + 19x + 21 = 0
4x² + 12x + 7x + 21 = 0,
So x = -3, – 1.75
2y² – 25y – 27 = 0
2y² – 27y + 2y – 27 = 0
So y = 13.5, -1

Incorrect

4x² + 19x + 21 = 0
4x² + 12x + 7x + 21 = 0,
So x = -3, – 1.75
2y² – 25y – 27 = 0
2y² – 27y + 2y – 27 = 0
So y = 13.5, -1

Unattempted

4x² + 19x + 21 = 0
4x² + 12x + 7x + 21 = 0,
So x = -3, – 1.75
2y² – 25y – 27 = 0
2y² – 27y + 2y – 27 = 0
So y = 13.5, -1
Question 30 of 35

30. Question
1 points
Directions (26-30): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

I. 6x² + x – 2 = 0,
II. 2y² + 11y + 14 = 0
- x = y or relation cannot be established
- x ≥ y
- x < y
- x ≤ y
- x > y
Correct

6x² + x – 2 = 0
6x² + 4x – 3x – 2 = 0
Gives x = -2/3, 1/2
2y² + 11y + 14 = 0
2y² + 4y + 7y + 14 = 0
Gives y = -7/2, -2

Incorrect

6x² + x – 2 = 0
6x² + 4x – 3x – 2 = 0
Gives x = -2/3, 1/2
2y² + 11y + 14 = 0
2y² + 4y + 7y + 14 = 0
Gives y = -7/2, -2

Unattempted

6x² + x – 2 = 0
6x² + 4x – 3x – 2 = 0
Gives x = -2/3, 1/2
2y² + 11y + 14 = 0
2y² + 4y + 7y + 14 = 0
Gives y = -7/2, -2
Question 31 of 35

31. Question
1 points
Directions (31-35): Study the following table carefully and answer the questions that follow:

What is the total number of students from State B who have taken part in Painting and from state D who have taken part in Indoor Play?
- 1170
- 1230
- 1190
- 1060
- 1050
Correct

(15/100)*1800 + (45/100)*2000

Incorrect

(15/100)*1800 + (45/100)*2000

Unattempted

(15/100)*1800 + (45/100)*2000
Question 32 of 35

32. Question
1 points
Directions (31-35): Study the following table carefully and answer the questions that follow:

Total number of students in Dance and Painting from State A is approximately what percent more than number of students participating in Cycling from State C?
- 10%
- 12%
- 14%
- 15%
- 18%
Correct

Dance and Painting from State A = (15/100)*2000 + (26/100)*1800 = 768
Cycling from State C = (28/100)*2400 = 672
Required% = (768-672)/672 *100

Incorrect

Dance and Painting from State A = (15/100)*2000 + (26/100)*1800 = 768
Cycling from State C = (28/100)*2400 = 672
Required% = (768-672)/672 *100

Unattempted

Dance and Painting from State A = (15/100)*2000 + (26/100)*1800 = 768
Cycling from State C = (28/100)*2400 = 672
Required% = (768-672)/672 *100
Question 33 of 35

33. Question
1 points
Directions (31-35): Study the following table carefully and answer the questions that follow:

What is the ratio of total number of students participating in dancing from States A and B to total number of students participating in singing from States B and C?
- 18 : 25
- 23 : 32
- 33 : 28
- 25 : 33
- None of these
Correct

dancing from States A and B = (15+35)/100 * 2000 = 1000
singing from States B and C = (25+35)/100 * 2200 = 1320
1000 : 1320

Incorrect

dancing from States A and B = (15+35)/100 * 2000 = 1000
singing from States B and C = (25+35)/100 * 2200 = 1320
1000 : 1320

Unattempted

dancing from States A and B = (15+35)/100 * 2000 = 1000
singing from States B and C = (25+35)/100 * 2200 = 1320
1000 : 1320
Question 34 of 35

34. Question
1 points
Directions (31-35): Study the following table carefully and answer the questions that follow:

If a total of 1200 students win from State B, then students participating in Cycling from state B are what percent of winners from State B?
- 34%
- 40%
- 42%
- 45%
- 32%
Correct

Cycling from state B = (20/100)*2400 = 480
Required % = (480/1200)*100

Incorrect

Cycling from state B = (20/100)*2400 = 480
Required % = (480/1200)*100

Unattempted

Cycling from state B = (20/100)*2400 = 480
Required % = (480/1200)*100
Question 35 of 35

35. Question
1 points
Directions (31-35): Study the following table carefully and answer the questions that follow:

What is the difference between number of students who took part in Singing in state D and Indoor Play from State A?
- 184
- 134
- 198
- 204
- 212
Correct

Singing in state D = (22/100)*2200 = 484
Indoor Play from State A = (15/100)*2000 = 300

Incorrect

Singing in state D = (22/100)*2200 = 484
Indoor Play from State A = (15/100)*2000 = 300

Unattempted

Singing in state D = (22/100)*2200 = 484
Indoor Play from State A = (15/100)*2000 = 300

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17 Thoughts to “Quant Sectional Test 2 for SBI PO 2017 Prelim Exam”

Laughing tym its nitrous oxide

March 11, 2017 at 3:06 pm

TQ 🙂
1. akduuuuuu
  
  March 11, 2017 at 3:36 pm
  
  palsssssssssssh :))
  1. Laughing tym its nitrous oxide
    
    March 11, 2017 at 3:40 pm
    
    OO- /
akduuuuuu

March 12, 2017 at 4:43 pm

mam 12th ques main kuch wrong hai plz check
akduuuuuu

March 12, 2017 at 4:51 pm

statement main B and C ki place pe A and C ka difference hona chahiye
1. Shubhra
  
  March 12, 2017 at 5:14 pm
  
  Oh yes. Typing mistake
  1. akduuuuuu
    
    March 12, 2017 at 5:19 pm
    
    ok mam
purvi

March 12, 2017 at 11:21 pm

thanku
dream girl ;;;;/@ :)

March 15, 2017 at 12:37 pm

Time has elapsed

You have reached 14 of 35 points, (40%)
Preeti

March 15, 2017 at 7:00 pm

thankuu mam
Ankit Saxena

March 23, 2017 at 12:29 am

Time has elapsed

You have reached 23 of 35 points, (65.71%)
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