Mock Test free SBI PO Prelims 2017. Quant Section Mock

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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
The owner of a shop raised the price of the table by x% and then subsequently decreased by x%. Due to this the price of table decreased by Rs 100. After some time, the same procedure was applied again after which the price of that table was Rs 2304. Find the original price of the table?
- Rs 2500
- Rs 2400
- Rs 2600
- Rs 1800
- Rs 2100
Correct

With successive formula:
loss% = x – x – x*x/100 = – x²/100
Let the original price be Rs p and after 1st step is Rs q
So p – q = 100
Also after first step, price q = (100 – x²/100)/100 * p = (10000- x²)/10000 * p ……… (1)
After some time, same process applied again of Rs q:
So
(10000- x²)/10000 * q = 2304
From (1) q/p = (10000- x²)/10000
So
q/p * q = 2304 or q² = p*2304
We have 2 equations
p – q = 100
and q² = p*2304
Solve, p = Rs 2500

Incorrect

With successive formula:
loss% = x – x – x*x/100 = – x²/100
Let the original price be Rs p and after 1st step is Rs q
So p – q = 100
Also after first step, price q = (100 – x²/100)/100 * p = (10000- x²)/10000 * p ……… (1)
After some time, same process applied again of Rs q:
So
(10000- x²)/10000 * q = 2304
From (1) q/p = (10000- x²)/10000
So
q/p * q = 2304 or q² = p*2304
We have 2 equations
p – q = 100
and q² = p*2304
Solve, p = Rs 2500

Unattempted

With successive formula:
loss% = x – x – x*x/100 = – x²/100
Let the original price be Rs p and after 1st step is Rs q
So p – q = 100
Also after first step, price q = (100 – x²/100)/100 * p = (10000- x²)/10000 * p ……… (1)
After some time, same process applied again of Rs q:
So
(10000- x²)/10000 * q = 2304
From (1) q/p = (10000- x²)/10000
So
q/p * q = 2304 or q² = p*2304
We have 2 equations
p – q = 100
and q² = p*2304
Solve, p = Rs 2500
Question 2 of 35

2. Question
1 points
Ratio of average age of A three years hence and B 1 year ago is 3 : 2. Sum of ages of A and B is 12 more than twice the age of C. Sum of ages of B and C is twice the age of A. Find the age of C.
- 21 years
- 33 years
- 34 years
- 25 years
- 26 years
Correct

(A+3)/(B-1) = 3/2
(A+B) = 2C + 12
B+C = 2A
Solve, C = 25 years

Incorrect

(A+3)/(B-1) = 3/2
(A+B) = 2C + 12
B+C = 2A
Solve, C = 25 years

Unattempted

(A+3)/(B-1) = 3/2
(A+B) = 2C + 12
B+C = 2A
Solve, C = 25 years
Question 3 of 35

3. Question
1 points
What is the probability of getting a sum of 7 on two dices thrown simultaneously?
- 1/5
- 1/4
- 1/3
- 1/6
- 1/2
Correct

Total pairs = 36
Pair for Sum 7 = (1,6), (6,1), (2, 5), (5,2), (3,4), (4,3)
So probability = 6/36 = 1/6

Incorrect

Total pairs = 36
Pair for Sum 7 = (1,6), (6,1), (2, 5), (5,2), (3,4), (4,3)
So probability = 6/36 = 1/6

Unattempted

Total pairs = 36
Pair for Sum 7 = (1,6), (6,1), (2, 5), (5,2), (3,4), (4,3)
So probability = 6/36 = 1/6
Question 4 of 35

4. Question
1 points
A and B started a business by investing Rs 2000 and Rs 1500 respectively. 4 months later A withdrew his whole money and C joined B by investing Rs 3000. If the difference between the shares of A and C together and B is Rs 3423, then what is the annual profit?
- Rs 12,725
- Rs 12,225
- Rs 15,275
- Rs 14,125
- Rs 13,755
Correct

Ratio of shares of A : B : C is
2000*4 : 1500*12 : 3000*8
4 : 9 : 12
[(12+4)-9]/(4+9+12) * x = 3423
Solve, x = Rs 12225

Incorrect

Ratio of shares of A : B : C is
2000*4 : 1500*12 : 3000*8
4 : 9 : 12
[(12+4)-9]/(4+9+12) * x = 3423
Solve, x = Rs 12225

Unattempted

Ratio of shares of A : B : C is
2000*4 : 1500*12 : 3000*8
4 : 9 : 12
[(12+4)-9]/(4+9+12) * x = 3423
Solve, x = Rs 12225
Question 5 of 35

5. Question
1 points
A boat can row at 6 kmph in still water. If the velocity of the stream be 2 kmph, then what is the time taken by the boat to go to a place 24 km upstream and back?
- 4 hours
- 9 hours
- 10 hours
- 7 hours
- 5 hours
Correct

Upstream speed = speed of boat – speed of stream=6 – 2 = 4
Downstream speed = speed of boat + speed of stream=6 + 2=8
Time taken to go upstream = distance/speed = 24/4 =6 hour
Time taken to go downstream = distance/speed =24/8 = 3 hour
Total time = 6+3 = 9 hour

Incorrect

Upstream speed = speed of boat – speed of stream=6 – 2 = 4
Downstream speed = speed of boat + speed of stream=6 + 2=8
Time taken to go upstream = distance/speed = 24/4 =6 hour
Time taken to go downstream = distance/speed =24/8 = 3 hour
Total time = 6+3 = 9 hour

Unattempted

Upstream speed = speed of boat – speed of stream=6 – 2 = 4
Downstream speed = speed of boat + speed of stream=6 + 2=8
Time taken to go upstream = distance/speed = 24/4 =6 hour
Time taken to go downstream = distance/speed =24/8 = 3 hour
Total time = 6+3 = 9 hour
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.)

40.19% of 555.09 + 15.89% of 375.01 = 15.14 × 17.02 + ?
- 37
- 22
- 18
- 27
- 33
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.)

41.93% of 549.99 + 18.98 × 26.88 = 32.09% of 220.04 + ?
- 674
- 632
- 679
- 645
- 656
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?
(Note: You are not expected to calculate the exact value.)

(64.993)² ÷ (423.05 ÷ 16.99) = 24.99 + ? ²
- 20
- 16
- 12
- 23
- 07
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.)

√790 ÷ 6.99 × 135.09 = 43.01 × 21.02 + ?
- 363
- -378
- 356
- -363
- 378
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.)

48.92 × 22.07 + 304.09 ÷ 19.08 = 653.99 + ?
- 420
- 450
- 410
- 480
- 440
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?

3, 5, 13, 49, 241, ?
- 1477
- 1441
- 1322
- 1285
- 1223
Correct

3*2 – 1, 5*3 – 2, 13*4 – 3, 49*5 – 4, 241*6 – 5

Incorrect

3*2 – 1, 5*3 – 2, 13*4 – 3, 49*5 – 4, 241*6 – 5

Unattempted

3*2 – 1, 5*3 – 2, 13*4 – 3, 49*5 – 4, 241*6 – 5
Question 12 of 35

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

580, 292, 148, 76, 40, ?
- 18
- 26
- 25
- 22
- 15
Correct

580/2 + 2, 292/2 + 2, 148/2 + 2, 76/2 + 2, 40/2 + 2,

Incorrect

580/2 + 2, 292/2 + 2, 148/2 + 2, 76/2 + 2, 40/2 + 2,

Unattempted

580/2 + 2, 292/2 + 2, 148/2 + 2, 76/2 + 2, 40/2 + 2,
Question 13 of 35

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

24, 20, 24, 50, ?, 1608.5
- 201
- 255
- 278
- 324
- 339
Correct

24*0.5 + 8
20*1 + 4
24*2 + 2
50*4 + 1
201*8 + 0.5

Incorrect

24*0.5 + 8
20*1 + 4
24*2 + 2
50*4 + 1
201*8 + 0.5

Unattempted

24*0.5 + 8
20*1 + 4
24*2 + 2
50*4 + 1
201*8 + 0.5
Question 14 of 35

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

68, 69, 60, 85, 36, ?
- 198
- 169
- 187
- 144
- 117
Correct

+ 1², – 3², + 5², – 7², + 9²

Incorrect

+ 1², – 3², + 5², – 7², + 9²

Unattempted

+ 1², – 3², + 5², – 7², + 9²
Question 15 of 35

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

2, 4, 7, 19, 56, ?
- 133
- 155
- 166
- 122
- 144
Correct

Double Difference Series

Incorrect

Double Difference Series

Unattempted

Double Difference Series
Question 16 of 35

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

Find the total number of units exported by country A if the number of units (above Rs 10 lacs) exported by country C is 12,160.
- 16,000
- 20,000
- 28,000
- 22,000
- 31,000
Correct

Let total units exported by all countries = x
38/100 * 16/100 * x = 12160
x = 200,000
so by country A = 11/100 * 200,000 = 22,000

Incorrect

Let total units exported by all countries = x
38/100 * 16/100 * x = 12160
x = 200,000
so by country A = 11/100 * 200,000 = 22,000

Unattempted

Let total units exported by all countries = x
38/100 * 16/100 * x = 12160
x = 200,000
so by country A = 11/100 * 200,000 = 22,000
Question 17 of 35

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

The ratio of the number of units (above Rs 10 lacs) exported by Country D to that of the number of units (above Rs 10 lacs) exported by Country A is
- 23 : 11
- 31 : 12
- 28 : 9
- 25 : 11
- 31 : 12
Correct

Let total number of units exported by all countries = x
So
(46/100) * (21/100) * x : (42/100) * (11/100) * x
23 : 11

Incorrect

Let total number of units exported by all countries = x
So
(46/100) * (21/100) * x : (42/100) * (11/100) * x
23 : 11

Unattempted

Let total number of units exported by all countries = x
So
(46/100) * (21/100) * x : (42/100) * (11/100) * x
23 : 11
Question 18 of 35

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

If the number of units exported by Country C is 32,000, then what is the number of units (above Rs 10 lacs) exported by country F?
- 13,760
- 11,440
- 14,920
- 12,100
- 15,100
Correct

Let total number of units exported by all countries = x
16/100 * x = 32,000.
So x = 2,00,000
Number of units (above Rs 10 lacs) exported by country F = (52/100) * (11/100) * 2,00,000 = 11440

Incorrect

Let total number of units exported by all countries = x
16/100 * x = 32,000.
So x = 2,00,000
Number of units (above Rs 10 lacs) exported by country F = (52/100) * (11/100) * 2,00,000 = 11440

Unattempted

Let total number of units exported by all countries = x
16/100 * x = 32,000.
So x = 2,00,000
Number of units (above Rs 10 lacs) exported by country F = (52/100) * (11/100) * 2,00,000 = 11440
Question 19 of 35

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

If number of units exported by countries F and G increase by 10% each and the number of units above Rs 10 lacs remains unchanged for all the countries, then find the number of units (above Rs 10 lacs) exported by Country G.
- 12,740
- 11,200
- 13,400
- 12,350
- Cannot be determined
Correct

Let total number of units exported by all countries = x
After 10% increase, units exported by country G = (110/100) * (10/100) * x
Number of units (above Rs 10 lacs) exported by country G = (58/100) * (110/100) * (10/100) * x
We don’t have the value of x, so cant be determined.

Incorrect

Let total number of units exported by all countries = x
After 10% increase, units exported by country G = (110/100) * (10/100) * x
Number of units (above Rs 10 lacs) exported by country G = (58/100) * (110/100) * (10/100) * x
We don’t have the value of x, so cant be determined.

Unattempted

Let total number of units exported by all countries = x
After 10% increase, units exported by country G = (110/100) * (10/100) * x
Number of units (above Rs 10 lacs) exported by country G = (58/100) * (110/100) * (10/100) * x
We don’t have the value of x, so cant be determined.
Question 20 of 35

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

If the number of units exported by Country E increases by 10% while that exported by country A reduces by 5% and the percentage of units above Rs 10 lacs remains same for all the countries, then find the ratio of the number of units (above Rs 10 lacs) exported by Country E to the number of units (above Rs 10 lacs) exported by Country A.
- 272 : 193
- 292 : 123
- 262 : 133
- 272 : 155
- 272 : 133
Correct

Let total number of units exported by all countries = x
After 10% increase, units exported by country E = (110/100) * (16/100) * x
After 5% decrease, units exported by country A = (95/100) * (11/100) * x
the ratio of the number of units (above Rs 10 lacs) exported by Country E to the number of units (above Rs 10 lacs) exported by Country A:
(51/100) * (110/100) * (16/100) * x : (42/100) * (95/100) * (11/100) * x

Incorrect

Let total number of units exported by all countries = x
After 10% increase, units exported by country E = (110/100) * (16/100) * x
After 5% decrease, units exported by country A = (95/100) * (11/100) * x
the ratio of the number of units (above Rs 10 lacs) exported by Country E to the number of units (above Rs 10 lacs) exported by Country A:
(51/100) * (110/100) * (16/100) * x : (42/100) * (95/100) * (11/100) * x

Unattempted

Let total number of units exported by all countries = x
After 10% increase, units exported by country E = (110/100) * (16/100) * x
After 5% decrease, units exported by country A = (95/100) * (11/100) * x
the ratio of the number of units (above Rs 10 lacs) exported by Country E to the number of units (above Rs 10 lacs) exported by Country A:
(51/100) * (110/100) * (16/100) * x : (42/100) * (95/100) * (11/100) * x
Question 21 of 35

21. Question
1 points
Directions (21-25): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

I. 3x² + 7x – 6 = 0
II. 6y² – y – 2 = 0
- x > y
- x ≤ y
- x < y
- x ≥ y
- x = y or relation cannot be established
Correct

3x² + 7x – 6 = 0
3x² + 9x – 2x – 6 = 0
Gives x = -3, 2/3
6y² – y – 2 = 0
6y² + 3y – 4y – 2 = 0
Gives y = -1/2, 2/3
Put on number line
-3… -1/2 … 2/3
When y = -1/2, y < x (2/3) and y > x (-3) – here relation cant be determined.

Incorrect

3x² + 7x – 6 = 0
3x² + 9x – 2x – 6 = 0
Gives x = -3, 2/3
6y² – y – 2 = 0
6y² + 3y – 4y – 2 = 0
Gives y = -1/2, 2/3
Put on number line
-3… -1/2 … 2/3
When y = -1/2, y < x (2/3) and y > x (-3) – here relation cant be determined.

Unattempted

3x² + 7x – 6 = 0
3x² + 9x – 2x – 6 = 0
Gives x = -3, 2/3
6y² – y – 2 = 0
6y² + 3y – 4y – 2 = 0
Gives y = -1/2, 2/3
Put on number line
-3… -1/2 … 2/3
When y = -1/2, y < x (2/3) and y > x (-3) – here relation cant be determined.
Question 22 of 35

22. Question
1 points
Directions (21-25): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

I. 6x² + 13x + 6 = 0
II. 2y² + 11y + 12 = 0
- x ≤ y
- x > y
- x ≥ y
- x = y or relation cannot be established
- x < y
Correct

6x² + 13x + 6 = 0
6x² + 9x + 4x + 6 = 0
Gives x = -2/3, -3/2
2y² + 11y + 12 = 0
2y² + 8y + 3y + 12 = 0
Gives y = -4, -3/2
Put on number line
-4… -3/2… -2/3

Incorrect

6x² + 13x + 6 = 0
6x² + 9x + 4x + 6 = 0
Gives x = -2/3, -3/2
2y² + 11y + 12 = 0
2y² + 8y + 3y + 12 = 0
Gives y = -4, -3/2
Put on number line
-4… -3/2… -2/3

Unattempted

6x² + 13x + 6 = 0
6x² + 9x + 4x + 6 = 0
Gives x = -2/3, -3/2
2y² + 11y + 12 = 0
2y² + 8y + 3y + 12 = 0
Gives y = -4, -3/2
Put on number line
-4… -3/2… -2/3
Question 23 of 35

23. Question
1 points
Directions (21-25): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

I. 6x² + 5x – 1 = 0
II. 3y² – 11y + 6 = 0
- x < y
- x = y or relation cannot be established
- x > y
- x ≤ y
- x ≥ y
Correct

6x² + 5x – 1 = 0
6x² + 6x – x – 1 = 0
Gives x = -1, 1/6
3y² – 11y + 6 = 0
3y² – 9y – 2y + 6 = 0
Gives y = 2/3, 3
Put on number line
-1… 1/6… 2/3… 3

Incorrect

6x² + 5x – 1 = 0
6x² + 6x – x – 1 = 0
Gives x = -1, 1/6
3y² – 11y + 6 = 0
3y² – 9y – 2y + 6 = 0
Gives y = 2/3, 3
Put on number line
-1… 1/6… 2/3… 3

Unattempted

6x² + 5x – 1 = 0
6x² + 6x – x – 1 = 0
Gives x = -1, 1/6
3y² – 11y + 6 = 0
3y² – 9y – 2y + 6 = 0
Gives y = 2/3, 3
Put on number line
-1… 1/6… 2/3… 3
Question 24 of 35

24. Question
1 points
Directions (21-25): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

I. 3x² + 14x – 5 = 0
II. 3y² – 19y + 6 = 0
- x < y
- x = y or relation cannot be established
- x ≥ y
- x > y
- x ≤ y
Correct

3x² + 14x – 5 = 0
3x² + 15x – x – 5 = 0
Gives x = -5, 1/3
3y² – 19y + 6 = 0
3y² – 18y – y + 6 = 0
Gives y = 1/3, 6
Put on number line
-5…. 1/3… 6

Incorrect

3x² + 14x – 5 = 0
3x² + 15x – x – 5 = 0
Gives x = -5, 1/3
3y² – 19y + 6 = 0
3y² – 18y – y + 6 = 0
Gives y = 1/3, 6
Put on number line
-5…. 1/3… 6

Unattempted

3x² + 14x – 5 = 0
3x² + 15x – x – 5 = 0
Gives x = -5, 1/3
3y² – 19y + 6 = 0
3y² – 18y – y + 6 = 0
Gives y = 1/3, 6
Put on number line
-5…. 1/3… 6
Question 25 of 35

25. Question
1 points
Directions (21-25): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

I. 4x² + 4x – 15 = 0
II. y² – 4y – 32 = 0
- x ≤ y
- x > y
- x = y or relation cannot be established
- x < y
- x ≥ y
Correct

4x² + 4x – 15 = 0
4x² + 10x – 6x – 15 = 0
Gives x = -5/2, 3/2
y² – 4y – 32 = 0
y² + 4y – 8y – 32 = 0
Gives y= -4, 8
Put on number line
-4…. -5/2… 3/2… 8

Incorrect

4x² + 4x – 15 = 0
4x² + 10x – 6x – 15 = 0
Gives x = -5/2, 3/2
y² – 4y – 32 = 0
y² + 4y – 8y – 32 = 0
Gives y= -4, 8
Put on number line
-4…. -5/2… 3/2… 8

Unattempted

4x² + 4x – 15 = 0
4x² + 10x – 6x – 15 = 0
Gives x = -5/2, 3/2
y² – 4y – 32 = 0
y² + 4y – 8y – 32 = 0
Gives y= -4, 8
Put on number line
-4…. -5/2… 3/2… 8
Question 26 of 35

26. Question
1 points
Directions (26–30) Study the given line graph carefully and answer the questions that follow.
The graph given below shows the number of students from different schools of city ABC taking part in painting and dancing for given period of time.

In which year is the percentage increase in the number of students over the previous year the highest for taking part in painting?
- 2010
- 2011
- 2012
- 2014
- 2015
Correct

In 2012, 2013 the number is decreasing
For other years it can be seen that in 2011, the increase is very high, so % will also be high as compared in other years.
It is not always applicable. Since the difference between 2600 and 1800 is much high then the difference between other years.

Incorrect

In 2012, 2013 the number is decreasing
For other years it can be seen that in 2011, the increase is very high, so % will also be high as compared in other years.
It is not always applicable. Since the difference between 2600 and 1800 is much high then the difference between other years.

Unattempted

In 2012, 2013 the number is decreasing
For other years it can be seen that in 2011, the increase is very high, so % will also be high as compared in other years.
It is not always applicable. Since the difference between 2600 and 1800 is much high then the difference between other years.
Question 27 of 35

27. Question
1 points
Directions (26–30) Study the given line graph carefully and answer the questions that follow.
The graph given below shows the number of students from different schools of city ABC taking part in painting and dancing for given period of time.

Number of students who took part in dancing is approximately what percentage of the total number of students taking part in both activities in the year 2012?
- 52%
- 32%
- 47%
- 44%
- 38%
Correct

In 2012, in dancing = 1900
In dancing + painting = 1900 + 2400 = 4300
So required % = (1900/4300)*100 = 44%

Incorrect

In 2012, in dancing = 1900
In dancing + painting = 1900 + 2400 = 4300
So required % = (1900/4300)*100 = 44%

Unattempted

In 2012, in dancing = 1900
In dancing + painting = 1900 + 2400 = 4300
So required % = (1900/4300)*100 = 44%
Question 28 of 35

28. Question
1 points
Directions (26–30) Study the given line graph carefully and answer the questions that follow.
The graph given below shows the number of students from different schools of city ABC taking part in painting and dancing for given period of time.

What is the average number of students who took part in paiting over all the years?
- 1900
- 2020
- 2130
- 1870
- 1800
Correct

(1400+2700+1900+2100+1200+1500)/6 = 1800

Incorrect

(1400+2700+1900+2100+1200+1500)/6 = 1800

Unattempted

(1400+2700+1900+2100+1200+1500)/6 = 1800
Question 29 of 35

29. Question
1 points
Directions (26–30) Study the given line graph carefully and answer the questions that follow.
The graph given below shows the number of students from different schools of city ABC taking part in painting and dancing for given period of time.

Number of students in dancing event in 2015 is what percent less than that in painting event in the same year?
- 22.3%
- 35.9%
- 31.8%
- 33.4%
- 28.8%
Correct

(2200-1500)/2200 * 100 = 31.8%

Incorrect

(2200-1500)/2200 * 100 = 31.8%

Unattempted

(2200-1500)/2200 * 100 = 31.8%
Question 30 of 35

30. Question
1 points
Directions (26–30) Study the given line graph carefully and answer the questions that follow.
The graph given below shows the number of students from different schools of city ABC taking part in painting and dancing for given period of time.

What is the difference between the averages of number of students in both events for the years 2010, 2011, 2012 and 2015?
- 375
- 280
- 336
- 302
- 295
Correct

Average in painting in given years= (1800+2600+2400+2200)/4 = 2250
Average in dancing in given years = 1400+2700+1900+1500 / 4 = 1875
Difference = 375

Incorrect

Average in painting in given years= (1800+2600+2400+2200)/4 = 2250
Average in dancing in given years = 1400+2700+1900+1500 / 4 = 1875
Difference = 375

Unattempted

Average in painting in given years= (1800+2600+2400+2200)/4 = 2250
Average in dancing in given years = 1400+2700+1900+1500 / 4 = 1875
Difference = 375
Question 31 of 35

31. Question
1 points
The perimeter of a rectangle of length 62 cm and breadth 50 cm is four times perimeter of a square. What will be the circumference of a semicircle whose diameter is equal to the side of the given square?
- 26 cm
- 24 cm
- 28 cm
- 22 cm
- 25 cm
Correct

Let the side of the square be a cm.
Parameter of the rectangle = 2(62 + 50) = 224 cm Parameter of the square = 56 cm
i.e. 4a = 56
So a = 14
Diameter, d of the semicircle = 14 cm
Circumference of the semicircle = 1/2(π)(r) + d
= 1/2(22/7)(7) + 14 = 25 cm

Incorrect

Let the side of the square be a cm.
Parameter of the rectangle = 2(62 + 50) = 224 cm Parameter of the square = 56 cm
i.e. 4a = 56
So a = 14
Diameter, d of the semicircle = 14 cm
Circumference of the semicircle = 1/2(π)(r) + d
= 1/2(22/7)(7) + 14 = 25 cm

Unattempted

Let the side of the square be a cm.
Parameter of the rectangle = 2(62 + 50) = 224 cm Parameter of the square = 56 cm
i.e. 4a = 56
So a = 14
Diameter, d of the semicircle = 14 cm
Circumference of the semicircle = 1/2(π)(r) + d
= 1/2(22/7)(7) + 14 = 25 cm
Question 32 of 35

32. Question
1 points
Sixteen men can complete a work in 12 days. Twenty four children can complete the same work in 18 days. Twelve men and eight children started working and after 8 days three more children joined them. How many days will they now take to complete the remaining work?
- 3
- 5
- 7
- 4
- None of these
Correct

16 m in 12 d, so 1 m in 16*12 days
24 c in 18 d, so 1 c in 24*18 days
So 12 m in 16*12/12 = 16 days and 8 c in 24*18/8 = 54 days
They work for 8 days, so work done by them is (1/16 + 1/54)*8 = 35/54 , remaining work is 1 – 35/54 = 19/54
Now 3 more children join, so 3 c work in 24*18/3 = 144 days
So (1/16 + 1/54 + 1/144)*x = 19/54
Solve, x = 4

Incorrect

16 m in 12 d, so 1 m in 16*12 days
24 c in 18 d, so 1 c in 24*18 days
So 12 m in 16*12/12 = 16 days and 8 c in 24*18/8 = 54 days
They work for 8 days, so work done by them is (1/16 + 1/54)*8 = 35/54 , remaining work is 1 – 35/54 = 19/54
Now 3 more children join, so 3 c work in 24*18/3 = 144 days
So (1/16 + 1/54 + 1/144)*x = 19/54
Solve, x = 4

Unattempted

16 m in 12 d, so 1 m in 16*12 days
24 c in 18 d, so 1 c in 24*18 days
So 12 m in 16*12/12 = 16 days and 8 c in 24*18/8 = 54 days
They work for 8 days, so work done by them is (1/16 + 1/54)*8 = 35/54 , remaining work is 1 – 35/54 = 19/54
Now 3 more children join, so 3 c work in 24*18/3 = 144 days
So (1/16 + 1/54 + 1/144)*x = 19/54
Solve, x = 4
Question 33 of 35

33. Question
1 points
A person lent out certain sum on simple interest and the same sum on compound interest at a certain rate of interest per annum. He noticed that the ratio between the difference of compound interest and simple interest of 3 years and 2 years is 51 : 16. The rate of interest per annum is
- 18 1/2%
- 16 5/6%
- 18 3/4%
- 19 2/3%
- 16 2/3%
Correct

For 2 years, diff in CI and SI = Pr²/100²
For 3 years diff is Pr²(r+300)/100³
Pr²/100² / Pr²(r+300)/100³
= 15/47
So 100/(r+300) = 16/51
Solve, r = 300/16 = 18 3/4%

Incorrect

For 2 years, diff in CI and SI = Pr²/100²
For 3 years diff is Pr²(r+300)/100³
Pr²/100² / Pr²(r+300)/100³
= 15/47
So 100/(r+300) = 16/51
Solve, r = 300/16 = 18 3/4%

Unattempted

For 2 years, diff in CI and SI = Pr²/100²
For 3 years diff is Pr²(r+300)/100³
Pr²/100² / Pr²(r+300)/100³
= 15/47
So 100/(r+300) = 16/51
Solve, r = 300/16 = 18 3/4%
Question 34 of 35

34. Question
1 points
There were 5000 students in a school. In a year school lost its reputation so 5% students left the school. The teachers work harder so the percentage of students grew by 5% and 10% in the subsequent two years. If these are nearby percentages, what is the approximate number of students in the school after 3 years?
- 5320
- 5490
- 5440
- 5390
- 5550
Correct

5000 (100 – 5/100) (100 + 5/100) (100 + 10/100)
= 5000 (95/100) (105/100) (110/100) = 95*105*11/20 = 5486.25 ≈ 5490

Incorrect

5000 (100 – 5/100) (100 + 5/100) (100 + 10/100)
= 5000 (95/100) (105/100) (110/100) = 95*105*11/20 = 5486.25 ≈ 5490

Unattempted

5000 (100 – 5/100) (100 + 5/100) (100 + 10/100)
= 5000 (95/100) (105/100) (110/100) = 95*105*11/20 = 5486.25 ≈ 5490
Question 35 of 35

35. Question
1 points
Average age of some members in a group is 56. The average decreases by 1 when 2 more members join the group. But if 2 members having same age as these leave the original group, the average age increases by 3. Find the initial number of members in the group.
- 4
- 10
- 6
- 5
- 11
Correct

Let initially x members in the group
Let sum of ages of x members = y
So
y/x = 56
Let sum of age of 2 members added = z
So (y+z)/(x+2) = 55
And when these leave
Then (y-z)/(x-2) = 58
Solve all three equations and find x, x= 6
OR
Average age of x nmembers is 56, of (x+2) is 55
So age of sum of these two members = 55*2 – x*1 = 110-x
Also
Average age of x members is 56, of (x-2) is 58
So age of sum of these two members = 58*2 – x*2 = 116-2x
So 110-x = 116-2x
Solve, x = 6

Incorrect

Let initially x members in the group
Let sum of ages of x members = y
So
y/x = 56
Let sum of age of 2 members added = z
So (y+z)/(x+2) = 55
And when these leave
Then (y-z)/(x-2) = 58
Solve all three equations and find x, x= 6
OR
Average age of x nmembers is 56, of (x+2) is 55
So age of sum of these two members = 55*2 – x*1 = 110-x
Also
Average age of x members is 56, of (x-2) is 58
So age of sum of these two members = 58*2 – x*2 = 116-2x
So 110-x = 116-2x
Solve, x = 6

Unattempted

Let initially x members in the group
Let sum of ages of x members = y
So
y/x = 56
Let sum of age of 2 members added = z
So (y+z)/(x+2) = 55
And when these leave
Then (y-z)/(x-2) = 58
Solve all three equations and find x, x= 6
OR
Average age of x nmembers is 56, of (x+2) is 55
So age of sum of these two members = 55*2 – x*1 = 110-x
Also
Average age of x members is 56, of (x-2) is 58
So age of sum of these two members = 58*2 – x*2 = 116-2x
So 110-x = 116-2x
Solve, x = 6

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10 Thoughts to “Quant Sectional Test 3 for SBI PO 2017 Prelim Exam”

ǟʍɮɨӄǟ

March 19, 2017 at 3:39 pm

math hi ni hota :(((

14 of 35 questions answered correctly

Your time: 00:23:52

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

Average score 0%
Your score 40%
Prachi Pandey

March 19, 2017 at 6:15 pm

please provide detailed solution of approximation question
1. Shubhra
  
  March 20, 2017 at 3:04 pm
  
  just take the round of
  
  See this
  40.19% of 555.09 + 15.89% of 375.01 = 15.14 × 17.02 + ?
  
  40% of 555 + 16% of 375 = 15*17 + ?
  find the required
  In exams also numbers like 34.99 or 45.08 comes – so just make them 35 and 45 respectively and solve
optimistic

March 20, 2017 at 2:50 pm

16,18,28 solns pls check
1. Shubhra
  
  March 20, 2017 at 3:10 pm
  
  Yes
  corrected
  ty
Ankit Saxena

March 22, 2017 at 7:00 pm

@Shubhra_AspirantsZone:disqus q-15 , in q-17 E in place of D Mam
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10 Thoughts to “Quant Sectional Test 3 for SBI PO 2017 Prelim Exam”