Free Mock Test SBI PO Prelims 2017 according to latest pattern

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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
A boat can row to a place 84 km away and come back in 26 hours. The time to row 21 km with the stream is same as the time to row 18 km against the stream. Find the speed of the stream.
- 6.5 km/hr
- 4.5 km/hr
- 0.5 km/hr
- 2.5 km/hr
- 1.5 km/hr
Correct

Downstream speed = 21/x km/hr
Upstream speed = 18/x km/hr
84/(21/x) + 84/(18/x) = 26
Solve, x = 3 km/hr
So, downstream speed = 7 km/hr, upstream speed = 6 km/hr
Speed of boat = 1/2 * (7 – 6) km/hr

Incorrect

Downstream speed = 21/x km/hr
Upstream speed = 18/x km/hr
84/(21/x) + 84/(18/x) = 26
Solve, x = 3 km/hr
So, downstream speed = 7 km/hr, upstream speed = 6 km/hr
Speed of boat = 1/2 * (7 – 6) km/hr

Unattempted

Downstream speed = 21/x km/hr
Upstream speed = 18/x km/hr
84/(21/x) + 84/(18/x) = 26
Solve, x = 3 km/hr
So, downstream speed = 7 km/hr, upstream speed = 6 km/hr
Speed of boat = 1/2 * (7 – 6) km/hr
Question 2 of 35

2. Question
1 points
If three dices are thrown simultaneously, what is the probability of having number 2 on third dice?
- 1/36
- 5/36
- 1/6
- 13/216
- 5/6
Correct

In all there sre 6^3 = 216 choices
On third dice 2 is fixed. On first 2 dices there can be any number
(1,1,2), (1,2,2)…..(1,6,2)
(2,1,2)…………..(2,6,2)
(3,1,2)…………..(3,6,2)

Like this, there are 36 favorable events
So probability = 36/216 = 1/6

Incorrect

In all there sre 6^3 = 216 choices
On third dice 2 is fixed. On first 2 dices there can be any number
(1,1,2), (1,2,2)…..(1,6,2)
(2,1,2)…………..(2,6,2)
(3,1,2)…………..(3,6,2)

Like this, there are 36 favorable events
So probability = 36/216 = 1/6

Unattempted

In all there sre 6^3 = 216 choices
On third dice 2 is fixed. On first 2 dices there can be any number
(1,1,2), (1,2,2)…..(1,6,2)
(2,1,2)…………..(2,6,2)
(3,1,2)…………..(3,6,2)

Like this, there are 36 favorable events
So probability = 36/216 = 1/6
Question 3 of 35

3. Question
1 points
6 years ago, three times the age of B was 2 more than the age if A that time. 6 years hence, twice age of B will be equal to A’s age that time. If the present age of C is the average of the present ages of A and B, find the age of C 5 years ago.
- 33 years
- 30 years
- 23 years
- 38 years
- 28 years
Correct

3 * (B-6) = 2 + (A-6)
2 * (B+6) = A + 6
Solve, A = 46, B = 20
So C’s present age = (46+20)/2 = 33
So 5 years ago = 28

Incorrect

3 * (B-6) = 2 + (A-6)
2 * (B+6) = A + 6
Solve, A = 46, B = 20
So C’s present age = (46+20)/2 = 33
So 5 years ago = 28

Unattempted

3 * (B-6) = 2 + (A-6)
2 * (B+6) = A + 6
Solve, A = 46, B = 20
So C’s present age = (46+20)/2 = 33
So 5 years ago = 28
Question 4 of 35

4. Question
1 points
Container A and B contains water and alcohol in the ratio 2 : 3 and 2 : 5 respectively. How much amount of mixture from container A should be mixed with 126 l of mixture from container B, so that the resultant mixture contains water and alcohol in the ratio 7 : 15?
- 52 l
- 60 l
- 58 l
- 50 l
- 63 l
Correct

Water in A = 2/5. Water in B = 2/7 . And in resultant water = 7/22
So by allegation method:
(x)…………….….(126)
2/5……………..…..2/7
………..7/22
5/(22*7)…………..9/(22*5)
Take ratio: 5/22*7 : 9/22*5
Gives 25 : 63
So x/126 = 25/63
Solve, x = 50 l

Incorrect

Water in A = 2/5. Water in B = 2/7 . And in resultant water = 7/22
So by allegation method:
(x)…………….….(126)
2/5……………..…..2/7
………..7/22
5/(22*7)…………..9/(22*5)
Take ratio: 5/22*7 : 9/22*5
Gives 25 : 63
So x/126 = 25/63
Solve, x = 50 l

Unattempted

Water in A = 2/5. Water in B = 2/7 . And in resultant water = 7/22
So by allegation method:
(x)…………….….(126)
2/5……………..…..2/7
………..7/22
5/(22*7)…………..9/(22*5)
Take ratio: 5/22*7 : 9/22*5
Gives 25 : 63
So x/126 = 25/63
Solve, x = 50 l
Question 5 of 35

5. Question
1 points
A and B start a business by investing Rs 2400 and Rs 2100 respectively. After 3 months A adds Rs 800 and B adds Rs 300 to their investments and third partner C joins them by investing Rs 3200. If share of B after a year is Rs 17,050 out of the total profit, find the profit of C.
- Rs 17,600
- Rs 16,400
- Rs 15,600
- Rs 18,200
- Rs 16,900
Correct

Ratio shares of A : B : C is
2400*3 + 3200*9 : 2100*3 + 2400*9 : 3200*9
24*3 + 32*9 : 21*3 + 24*9 : 32*9
8 + 32 : 7 + 24 : 32
40 : 31 : 32
So
31/32 = 17050/x
Solve, x = Rs 17,600

Incorrect

Ratio shares of A : B : C is
2400*3 + 3200*9 : 2100*3 + 2400*9 : 3200*9
24*3 + 32*9 : 21*3 + 24*9 : 32*9
8 + 32 : 7 + 24 : 32
40 : 31 : 32
So
31/32 = 17050/x
Solve, x = Rs 17,600

Unattempted

Ratio shares of A : B : C is
2400*3 + 3200*9 : 2100*3 + 2400*9 : 3200*9
24*3 + 32*9 : 21*3 + 24*9 : 32*9
8 + 32 : 7 + 24 : 32
40 : 31 : 32
So
31/32 = 17050/x
Solve, x = Rs 17,600
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?
(Note: You are not expected to calculate the exact value.)

20.96% of 235.90 + 54.04% of 149.98 – 3.96% of 102.05 = ?
- 136
- 127
- 102
- 145
- 153
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?
(Note: You are not expected to calculate the exact value.)

17.99² – 14.05² + (2350.05+80.25) ÷ ? = 230
- 18
- 14
- 38
- 33
- 24
Correct

17.99² – 14.05² + (2350.05+80.05) ÷ ? = 230
18² – 14² + (2350+80) ÷ ? = 230
130 + (2430) ÷ ? = 230
? = 24.3

Incorrect

17.99² – 14.05² + (2350.05+80.05) ÷ ? = 230
18² – 14² + (2350+80) ÷ ? = 230
130 + (2430) ÷ ? = 230
? = 24.3

Unattempted

17.99² – 14.05² + (2350.05+80.05) ÷ ? = 230
18² – 14² + (2350+80) ÷ ? = 230
130 + (2430) ÷ ? = 230
? = 24.3
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?
(Note: You are not expected to calculate the exact value.)

(3/4) + (2/3) + (1 1/3) + (2 2/3) = ?
- 10
- 34
- 25
- 5
- 43
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?
(Note: You are not expected to calculate the exact value.)

25.09 × 6.9 + 25.92% of 220.09 – √492 = ?
- 213
- 201
- 184
- 229
- 168
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?
(Note: You are not expected to calculate the exact value.)

45.009% of 640.08 – ? % of 959.99 = 48.06
- 34
- 39
- 18
- 25
- 42
Correct

Incorrect

Unattempted
Question 11 of 35

11. Question
1 points
Directions (11-15): What will come in place of question mark (?) in the following number series?

4, 5, 13, 23, 49, ?
- 95
- 106
- 87
- 145
- 166
Correct

×2 – 3, ×2 + 3, ×2 – 3, ×2 + 3, ….

Incorrect

×2 – 3, ×2 + 3, ×2 – 3, ×2 + 3, ….

Unattempted

×2 – 3, ×2 + 3, ×2 – 3, ×2 + 3, ….
Question 12 of 35

12. Question
1 points
Directions (11-15): What will come in place of question mark (?) in the following number series?

18, 9, 13.5, 33.75, ?
- 109.225
- 136.725
- 120.525
- 118.125
- 130.125
Correct

18*0.5, 9*1.5, 13.5*2.5, 33.75*3.5,

Incorrect

18*0.5, 9*1.5, 13.5*2.5, 33.75*3.5,

Unattempted

18*0.5, 9*1.5, 13.5*2.5, 33.75*3.5,
Question 13 of 35

13. Question
1 points
Directions (11-15): What will come in place of question mark (?) in the following number series?

20, 19, 36, 105, 416, ?
- 1525
- 2612
- 1865
- 2075
- 1022
Correct

20*1 – 1, 19*2 – 2, 36*3 – 3, 105*4 – 4,

Incorrect

20*1 – 1, 19*2 – 2, 36*3 – 3, 105*4 – 4,

Unattempted

20*1 – 1, 19*2 – 2, 36*3 – 3, 105*4 – 4,
Question 14 of 35

14. Question
1 points
Directions (11-15): What will come in place of question mark (?) in the following number series?

13, 17, 24, 37, 59, ?
- 102
- 98
- 93
- 121
- 126
Correct

Double difference series
17 – 13 =4, 24 – 17 = 7, 37 – 24 = 13, 59-37 = 22
7 – 4 = 3, 13 – 7 = 6, 22 – 13 = 9
So next number 12+22+59

Incorrect

Double difference series
17 – 13 =4, 24 – 17 = 7, 37 – 24 = 13, 59-37 = 22
7 – 4 = 3, 13 – 7 = 6, 22 – 13 = 9
So next number 12+22+59

Unattempted

Double difference series
17 – 13 =4, 24 – 17 = 7, 37 – 24 = 13, 59-37 = 22
7 – 4 = 3, 13 – 7 = 6, 22 – 13 = 9
So next number 12+22+59
Question 15 of 35

15. Question
1 points
Directions (11-15): What will come in place of question mark (?) in the following number series?

7, 12, 27, 72, 207, ?
- 566
- 612
- 645
- 422
- 534
Correct

12 – 7 = 5, 27 – 12 = 15, 72 – 27 = 45, 207 – 72 = 135
5*3⁰, 5*3¹, 5*3², 5*3³, 5*3⁴

Incorrect

12 – 7 = 5, 27 – 12 = 15, 72 – 27 = 45, 207 – 72 = 135
5*3⁰, 5*3¹, 5*3², 5*3³, 5*3⁴

Unattempted

12 – 7 = 5, 27 – 12 = 15, 72 – 27 = 45, 207 – 72 = 135
5*3⁰, 5*3¹, 5*3², 5*3³, 5*3⁴
Question 16 of 35

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

How many marks did Kavya get in all the subjects?
- 456
- 463
- 471
- 474
- 422
Correct

Total marks of Kavya = 66/100 * 150 + 75 + 88/100 * 150 + 56/100 * 125 + 56/100 * 75 + 45

Incorrect

Total marks of Kavya = 66/100 * 150 + 75 + 88/100 * 150 + 56/100 * 125 + 56/100 * 75 + 45

Unattempted

Total marks of Kavya = 66/100 * 150 + 75 + 88/100 * 150 + 56/100 * 125 + 56/100 * 75 + 45
Question 17 of 35

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

Marks obtained by Medha in Hindi is what percent of marks obtained by Sheena in the same subject?
- 115.8%
- 123.8%
- 175.5%
- 96.2%
- None of these
Correct

Required % = 88/76 * 100 = 115.8%

Incorrect

Required % = 88/76 * 100 = 115.8%

Unattempted

Required % = 88/76 * 100 = 115.8%
Question 18 of 35

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

what are the average marks obtained by all students together in Computer?
- 123
- 137
- 116
- 109
- 128
Correct

Average marks in Computer = (88+84+78+96+68+50)/6 = 464/6%

Required average marks = 464/600 * 150 = 116

Incorrect

Average marks in Computer = (88+84+78+96+68+50)/6 = 464/6%

Required average marks = 464/600 * 150 = 116

Unattempted

Average marks in Computer = (88+84+78+96+68+50)/6 = 464/6%

Required average marks = 464/600 * 150 = 116
Question 19 of 35

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

Who has scored the highest total marks in all the subjects together?
- Kavya
- Garima
- Medha
- Shruti
- Sheena
Correct

Total by Medha = 568 – the highest

Incorrect

Total by Medha = 568 – the highest

Unattempted

Total by Medha = 568 – the highest
Question 20 of 35

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

How many students have scored the highest marks in more than one subject?
- None
- One
- Two
- Three
- More than three
Correct

Sheena in Physics and Chemistry

Medha in English and Computer

Incorrect

Sheena in Physics and Chemistry

Medha in English and Computer

Unattempted

Sheena in Physics and Chemistry

Medha in English and Computer
Question 21 of 35

21. Question
1 points
Directions (21-25): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-

I. 20x² – 31x + 12 = 0,
II. 6y² – 7y + 2 = 0
- If x < y
- If x > y
- If x ≥ y
- If x = y or relation cannot be established
- If x ≤ y
Correct

20x² – 31x + 12 = 0
20x² – 16x – 15x + 12 = 0
So x = 3/4, 4/5
6y² – 7y + 2 = 0
6y² – 3y – 4y + 2 = 0
So y = 1/2, 2/3

Incorrect

20x² – 31x + 12 = 0
20x² – 16x – 15x + 12 = 0
So x = 3/4, 4/5
6y² – 7y + 2 = 0
6y² – 3y – 4y + 2 = 0
So y = 1/2, 2/3

Unattempted

20x² – 31x + 12 = 0
20x² – 16x – 15x + 12 = 0
So x = 3/4, 4/5
6y² – 7y + 2 = 0
6y² – 3y – 4y + 2 = 0
So y = 1/2, 2/3
Question 22 of 35

22. Question
1 points
Directions (21-25): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-

I. 3x² + 22 x + 24 = 0,
II. 3y² – 10y + 3 = 0
- If x ≥ y
- If x > y
- If x ≤ y
- If x < y
- If x = y or relation cannot be established
Correct

3x² + 22 x + 24 = 0
3x² + 18x + 4x + 24 = 0
So x = -4/3, -6
3y² – 10y + 3 = 0
3y² – 9y – y + 3 = 0
So y = 1/3, 3

Incorrect

3x² + 22 x + 24 = 0
3x² + 18x + 4x + 24 = 0
So x = -4/3, -6
3y² – 10y + 3 = 0
3y² – 9y – y + 3 = 0
So y = 1/3, 3

Unattempted

3x² + 22 x + 24 = 0
3x² + 18x + 4x + 24 = 0
So x = -4/3, -6
3y² – 10y + 3 = 0
3y² – 9y – y + 3 = 0
So y = 1/3, 3
Question 23 of 35

23. Question
1 points
Directions (21-25): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-

I. x² – x – 6 = 0,
II. 5y² – 7y – 6 = 0
- If x ≤ y
- If x > y
- E) If x = y or relation cannot be established
- If x < y
- If x ≥ y
Correct

x² – x – 6 = 0
x² – 2x + 3x – 6 = 0
So x = -3, 2
5y² – 7y – 6 = 0
5y² – 10y + 3y – 6 = 0
So y = -3/5, 2

Incorrect

x² – x – 6 = 0
x² – 2x + 3x – 6 = 0
So x = -3, 2
5y² – 7y – 6 = 0
5y² – 10y + 3y – 6 = 0
So y = -3/5, 2

Unattempted

x² – x – 6 = 0
x² – 2x + 3x – 6 = 0
So x = -3, 2
5y² – 7y – 6 = 0
5y² – 10y + 3y – 6 = 0
So y = -3/5, 2
Question 24 of 35

24. Question
1 points
Directions (21-25): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-

I. 2x² + 17x + 30 = 0,
II. 2y² + 13y + 18 = 0
- If x ≤ y
- If x = y or relation cannot be established
- If x > y
- If x ≥ y
- If x < y
Correct

2x² + 17x + 30 = 0
2x² + 12x + 5x + 30 = 0
So x = -6, -5/2
2y² + 13y + 18 = 0
2y² + 4y + 9y + 18 = 0
So y = -9/2, -2

Incorrect

2x² + 17x + 30 = 0
2x² + 12x + 5x + 30 = 0
So x = -6, -5/2
2y² + 13y + 18 = 0
2y² + 4y + 9y + 18 = 0
So y = -9/2, -2

Unattempted

2x² + 17x + 30 = 0
2x² + 12x + 5x + 30 = 0
So x = -6, -5/2
2y² + 13y + 18 = 0
2y² + 4y + 9y + 18 = 0
So y = -9/2, -2
Question 25 of 35

25. Question
1 points
Directions (21-25): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-

I. 3x² + 16x + 20 = 0,
II. 3y² + 8y + 4 = 0
- If x = y or relation cannot be established
- If x < y
- If x ≥ y
- If x > y
- If x ≤ y
Correct

3x² + 16x + 20 = 0
3x² + 6x + 10x + 20 = 0
So x = -10/3, -2
3y² + 8y + 4 = 0
3y² + 6y + 2y + 4 = 0
So y = -2, -2/3

Incorrect

3x² + 16x + 20 = 0
3x² + 6x + 10x + 20 = 0
So x = -10/3, -2
3y² + 8y + 4 = 0
3y² + 6y + 2y + 4 = 0
So y = -2, -2/3

Unattempted

3x² + 16x + 20 = 0
3x² + 6x + 10x + 20 = 0
So x = -10/3, -2
3y² + 8y + 4 = 0
3y² + 6y + 2y + 4 = 0
So y = -2, -2/3
Question 26 of 35

26. Question
1 points
Directions (26-30): Study the following charts carefully to answer the questions that follow:

The pie chart shows the distribution of number of phones sold by a shop during different months.The table shows the respective ratio between the number of phones sold by shop of company A and company B in different months.

What is the respective ratio between the number of phones sold of company B during January and those sold during June of the same company?
- 123 : 135
- 119 : 139
- 119 : 123
- 119 : 135
- None of these
Correct

Ratio is
17*7/15 : 9*16/16
17*7 : 9*15
119 : 135

Incorrect

Ratio is
17*7/15 : 9*16/16
17*7 : 9*15
119 : 135

Unattempted

Ratio is
17*7/15 : 9*16/16
17*7 : 9*15
119 : 135
Question 27 of 35

27. Question
1 points
Directions (26-30): Study the following charts carefully to answer the questions that follow:

The pie chart shows the distribution of number of phones sold by a shop during different months.The table shows the respective ratio between the number of phones sold by shop of company A and company B in different months.

If 35% of the phones sold by company A during May were sold at a discount, home many phones of company A that month were sold without a discount?
- 1657
- 1638
- 1569
- 1602
- 1523
Correct

Phones sold by company A during May = 7/15 * 12/100 * 45000
So sold without discount = 65/100 * 7/15 * 12/100 * 45000 = 1638

Incorrect

Phones sold by company A during May = 7/15 * 12/100 * 45000
So sold without discount = 65/100 * 7/15 * 12/100 * 45000 = 1638

Unattempted

Phones sold by company A during May = 7/15 * 12/100 * 45000
So sold without discount = 65/100 * 7/15 * 12/100 * 45000 = 1638
Question 28 of 35

28. Question
1 points
Directions (26-30): Study the following charts carefully to answer the questions that follow:

The pie chart shows the distribution of number of phones sold by a shop during different months.The table shows the respective ratio between the number of phones sold by shop of company A and company B in different months.

If the shopkeeper earned a profit of Rs 420 on each phone sold of company B during April, what was his total profit earned on the phones of that company during the same month?
- Rs 6,30,000
- Rs 6,40,000
- Rs 6,20,000
- Rs 6,10,000
- Rs 6,50,000
Correct

Phones sold by B during April = 5/12 * 8/100 * 45000 = 1500
So profit earned = 420*1500 = Rs 6,30,000

Incorrect

Phones sold by B during April = 5/12 * 8/100 * 45000 = 1500
So profit earned = 420*1500 = Rs 6,30,000

Unattempted

Phones sold by B during April = 5/12 * 8/100 * 45000 = 1500
So profit earned = 420*1500 = Rs 6,30,000
Question 29 of 35

29. Question
1 points
Directions (26-30): Study the following charts carefully to answer the questions that follow:

The pie chart shows the distribution of number of phones sold by a shop during different months.The table shows the respective ratio between the number of phones sold by shop of company A and company B in different months.

The number of phones sold by company A during January is approximately what percent of the number of phones sold by company A during June?
- 98%
- 108%
- 130%
- 124%
- 140%
Correct

Sold by A in January = 8/15 * 17/100 * 45000 = 4080
in June = 7/16 * 16/100 * 45000 = 3150
Required % = 4080/3150 * 100 = 130%

Incorrect

Sold by A in January = 8/15 * 17/100 * 45000 = 4080
in June = 7/16 * 16/100 * 45000 = 3150
Required % = 4080/3150 * 100 = 130%

Unattempted

Sold by A in January = 8/15 * 17/100 * 45000 = 4080
in June = 7/16 * 16/100 * 45000 = 3150
Required % = 4080/3150 * 100 = 130%
Question 30 of 35

30. Question
1 points
Directions (26-30): Study the following charts carefully to answer the questions that follow:

The pie chart shows the distribution of number of phones sold by a shop during different months.The table shows the respective ratio between the number of phones sold by shop of company A and company B in different months.

What is the total number of phones sold of company B during February and March together?
- 12,000
- 9,000
- 11,500
- 9,500
- 10,000
Correct

Sold by B in Feb = 5/9 * 22/100 * 45000 = 5500
Sold by B in March = 2/5 * 25/100 * 45000 = 45000
So total = 10,000

Incorrect

Sold by B in Feb = 5/9 * 22/100 * 45000 = 5500
Sold by B in March = 2/5 * 25/100 * 45000 = 45000
So total = 10,000

Unattempted

Sold by B in Feb = 5/9 * 22/100 * 45000 = 5500
Sold by B in March = 2/5 * 25/100 * 45000 = 45000
So total = 10,000
Question 31 of 35

31. Question
1 points
A work which is completed by 20 men in 8 days can be completed by 25 women 12 days. 16 men and 10 women start doing the work. After 3 days, they leave. If the remaining work is to be completed in 6 days by x number of men, find x.
- 13
- 15
- 16
- 18
- 21
Correct

20 men in 8 days so 16 men in 20 × 8/16 = 10 days and
25 women in 12 days so 10 women in 25 × 12/10 = 30 days
So in 3 days, they complete (1/10 + 1/30) × 3 = 2/5
So remaining work = 1 – 2/5 = 3/5
20 m 1 work in 8 days and x men 3/5 work in 6 days
So 20 × 8 × 3/5 = x × 6 × 1
So, x = 16 men

Incorrect

20 men in 8 days so 16 men in 20 × 8/16 = 10 days and
25 women in 12 days so 10 women in 25 × 12/10 = 30 days
So in 3 days, they complete (1/10 + 1/30) × 3 = 2/5
So remaining work = 1 – 2/5 = 3/5
20 m 1 work in 8 days and x men 3/5 work in 6 days
So 20 × 8 × 3/5 = x × 6 × 1
So, x = 16 men

Unattempted

20 men in 8 days so 16 men in 20 × 8/16 = 10 days and
25 women in 12 days so 10 women in 25 × 12/10 = 30 days
So in 3 days, they complete (1/10 + 1/30) × 3 = 2/5
So remaining work = 1 – 2/5 = 3/5
20 m 1 work in 8 days and x men 3/5 work in 6 days
So 20 × 8 × 3/5 = x × 6 × 1
So, x = 16 men
Question 32 of 35

32. Question
1 points
The circumference of a circle having radius equal to 35 cm is equal to the perimeter of a rectangle. If the area of rectangle is 2400 cm², find the length of rectangle.
- 70 cm
- 80 cm
- 60 cm
- 55 cm
- 75 cm
Correct

2 × 22/7 × 35 = 2 (l + b)
so (l + b) = 110
also given, lb = 2400
So (l + 2400/l) = 110
So l² – 110 l + 2400 = 0
So, l = 80 or 30. 30 not present in options.

Incorrect

2 × 22/7 × 35 = 2 (l + b)
so (l + b) = 110
also given, lb = 2400
So (l + 2400/l) = 110
So l² – 110 l + 2400 = 0
So, l = 80 or 30. 30 not present in options.

Unattempted

2 × 22/7 × 35 = 2 (l + b)
so (l + b) = 110
also given, lb = 2400
So (l + 2400/l) = 110
So l² – 110 l + 2400 = 0
So, l = 80 or 30. 30 not present in options.
Question 33 of 35

33. Question
1 points
A trader gets a profit of 30% on buying a kg of wheat. By mistake, he sold 1200 grams of wheat at the price of 1 kg. Find his net profit/loss%.
- 55/3% profit
- 25/3% profit
- 125/3% loss
- 30/11% loss
- 25/6% profit
Correct

He sold 1200 grams of wheat at the price of 1 kg, means he incurs a loss here.
That loss% = (1200-1000)/1200 * 100 = 50/3%
Now there was a profit of 30% while buying and now a loss of 50/3% while selling. So successive method can be used here:
Profit/loss% = 30 + (-50/3) + (30)(-50/3)/100 = + 25/3%

Incorrect

He sold 1200 grams of wheat at the price of 1 kg, means he incurs a loss here.
That loss% = (1200-1000)/1200 * 100 = 50/3%
Now there was a profit of 30% while buying and now a loss of 50/3% while selling. So successive method can be used here:
Profit/loss% = 30 + (-50/3) + (30)(-50/3)/100 = + 25/3%

Unattempted

He sold 1200 grams of wheat at the price of 1 kg, means he incurs a loss here.
That loss% = (1200-1000)/1200 * 100 = 50/3%
Now there was a profit of 30% while buying and now a loss of 50/3% while selling. So successive method can be used here:
Profit/loss% = 30 + (-50/3) + (30)(-50/3)/100 = + 25/3%
Question 34 of 35

34. Question
1 points
In a class, average age of 30 students is 18 years. If the age of 2 more students is taken into consideration, then the average of all students gets increase by 1. Find the average of the ages of those 2 students.
- 38
- 26
- 21
- 34
- 42
Correct

30 students – 18
32 students – 19
So total age of those 2 students = 30×1 + 19×2 = 68
So average = 68/2 = 34

Incorrect

30 students – 18
32 students – 19
So total age of those 2 students = 30×1 + 19×2 = 68
So average = 68/2 = 34

Unattempted

30 students – 18
32 students – 19
So total age of those 2 students = 30×1 + 19×2 = 68
So average = 68/2 = 34
Question 35 of 35

35. Question
1 points
A person invests Rs 5800 in two schemes in such a way that he after 5 years and 4 years, he gets the same interest on respective parts. If the schemes offer 12% and 14% rate of interest respectively, what is the amount invested at 12% rate of interest?
- Rs 2600
- Rs 3100
- Rs 2300
- Rs 3600
- Rs 2800
Correct

Shortcut when interests are equal after t1 and t2 years:
1/(t1*r1) : 1/(t2*r2)
1/(5*12) : 1/(4*14)
14 : 15
So amount invested at 12% rate of interest = (14/29) * 5800 = 2800

Incorrect

Shortcut when interests are equal after t1 and t2 years:
1/(t1*r1) : 1/(t2*r2)
1/(5*12) : 1/(4*14)
14 : 15
So amount invested at 12% rate of interest = (14/29) * 5800 = 2800

Unattempted

Shortcut when interests are equal after t1 and t2 years:
1/(t1*r1) : 1/(t2*r2)
1/(5*12) : 1/(4*14)
14 : 15
So amount invested at 12% rate of interest = (14/29) * 5800 = 2800

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25 Thoughts to “Quant Sectional Test 5 for SBI PO 2017 Prelim Exam”

Di Ghosh

April 2, 2017 at 8:17 pm

31 recheck pls
1. EKTA
  
  April 3, 2017 at 1:35 pm
  
  crct he
Di Ghosh

April 2, 2017 at 8:30 pm

thanks
•?((¯°·._.• ąʍβɨЌą •._.·°¯))؟•

April 3, 2017 at 8:11 pm

18 of 35 questions answered correctly

Your time: 00:23:41

You have reached 18 of 35 points, (51.43)
Ravi kiran

April 4, 2017 at 12:05 pm

Madam March month current affairs in PDF format
Ravi kiran

April 4, 2017 at 12:08 pm

@shubhra madam
1. Shubhra
  
  April 4, 2017 at 1:57 pm
  
  Yes check in downloads section in menu
jaga

April 6, 2017 at 11:14 pm

NO4………….How much amount of mixture from container A should be mixed with 126 l of mixture from container B…ye line samjh mai nahi araha..
1. Shubhra
  
  April 7, 2017 at 8:56 am
  
  Container A me se kitna amount , container B k 126 l k sath mix krna chaiye?
  1. jaga
    
    April 7, 2017 at 9:42 am
    
    Hn
  2. jaga
    
    April 7, 2017 at 9:42 am
    
    Thqnk u
sachin shukla@Target sbi po

April 7, 2017 at 8:13 am

sir give me all concepts of RANKING test reasoning section ka?? plzzzzzz
MITTHU

April 7, 2017 at 5:10 pm

19 of 35 questions answered correctly

Your time: 00:24:00
GiGi

April 7, 2017 at 10:05 pm

Thanks alot ?
Prachi Pandey

April 8, 2017 at 8:51 pm

explain q no 7 please
1. Shubhra
  
  April 9, 2017 at 12:22 am
  
  18^2 – 14^2 = 128
  
  (2350+80)/x = 2430/x
  
  so 128 + 2430/x = 230
  x = 2430/102 = 24
smile

April 9, 2017 at 10:25 am

Hlo frnds !! Is der any free full length mock test for SBI po,if der.pls send me dat link.my mail id swathimailapalli221995@gmail.com.thanks in advance..
1. Shubhra
  
  April 9, 2017 at 11:15 am
  
  We will now add some full length mock test as SBI PO pre is near now
  Check here tomorrow
  1. LAUGHYNG TYM ITS NITROUS OXIDE
    
    April 9, 2017 at 11:22 am
    
    mam some quadratic equations sums comprises of huge numbers andd factoring them means break them into prime and consume them takes lots of time is there any otherway for solving it
    1. Shubhra
      
      April 9, 2017 at 11:31 am
      
      Yes there is a formula for calculating roots.
      Suppose equation is ax^2 + bx + c
      So its roots are [-b +- root (b^2 – 4ac)]/2a
      
      +- means ek root k liye plus hoga, and second k liye minus.
      
      But exam me se 5 questions me se 4 questions ki equations me jada bade factors nhi aate. To 5 me se 4 b attempt karoge to ok hai. Kabi kabi to all 5 are easy.
      
      Is formula se dekho jaldi solve hota hai to use it, otherwise leave
      1. LAUGHYNG TYM ITS NITROUS OXIDE
        
        April 9, 2017 at 11:33 am
        
        and mam have problem in like
        1+x/4 like that some pllz provide quises on that time and is they come in exam like instalment i never hear it come
      2. Shubhra
        
        April 9, 2017 at 3:04 pm
        
        Yes installment k kabi nhi aae phle.
        
        N 1+x/4 means ?
        I dint get
MiMi

April 14, 2017 at 12:13 am

Results
16 of 35 questions answered correctly

Your time: 00:24:00
dream girl ;;;;/@ :)

April 16, 2017 at 12:08 pm

Time has elapsed

You have reached 14 of 35 points, (40%)
el test de personalidad de meetic

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