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?
Correct
With successive formula:
loss% = x – x – x*x/100 = – x^{2}/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 – x^{2}/100)/100 * p = (10000- x^{2})/10000 * p ……… (1)
After some time, same process applied again of Rs q:
So
(10000- x^{2})/10000 * q = 2304
From (1) q/p = (10000- x^{2})/10000
So
q/p * q = 2304 or q^{2} = p*2304
We have 2 equations
p – q = 100
and q^{2} = p*2304
Solve, p = Rs 2500
Incorrect
With successive formula:
loss% = x – x – x*x/100 = – x^{2}/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 – x^{2}/100)/100 * p = (10000- x^{2})/10000 * p ……… (1)
After some time, same process applied again of Rs q:
So
(10000- x^{2})/10000 * q = 2304
From (1) q/p = (10000- x^{2})/10000
So
q/p * q = 2304 or q^{2} = p*2304
We have 2 equations
p – q = 100
and q^{2} = p*2304
Solve, p = Rs 2500
Unattempted
With successive formula:
loss% = x – x – x*x/100 = – x^{2}/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 – x^{2}/100)/100 * p = (10000- x^{2})/10000 * p ……… (1)
After some time, same process applied again of Rs q:
So
(10000- x^{2})/10000 * q = 2304
From (1) q/p = (10000- x^{2})/10000
So
q/p * q = 2304 or q^{2} = p*2304
We have 2 equations
p – q = 100
and q^{2} = 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.
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?
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?
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?
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 + ?
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 + ?
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}
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 + ?
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 + ?
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, ?
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?
Directions (11-15): What will come in place of question mark (?) in the following number series?
24, 20, 24, 50, ?, 1608.5
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, ?
Correct
+ 1^{2}, – 3^{2}, + 5^{2}, – 7^{2}, + 9^{2}
Incorrect
+ 1^{2}, – 3^{2}, + 5^{2}, – 7^{2}, + 9^{2}
Unattempted
+ 1^{2}, – 3^{2}, + 5^{2}, – 7^{2}, + 9^{2}
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, ?
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.
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
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?
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.
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.
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. 3x^{2} + 7x – 6 = 0
II. 6y^{2} – y – 2 = 0
Correct
3x^{2} + 7x – 6 = 0
3x^{2} + 9x – 2x – 6 = 0
Gives x = -3, 2/3
6y^{2} – y – 2 = 0
6y^{2} + 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
3x^{2} + 7x – 6 = 0
3x^{2} + 9x – 2x – 6 = 0
Gives x = -3, 2/3
6y^{2} – y – 2 = 0
6y^{2} + 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
3x^{2} + 7x – 6 = 0
3x^{2} + 9x – 2x – 6 = 0
Gives x = -3, 2/3
6y^{2} – y – 2 = 0
6y^{2} + 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. 6x^{2} + 13x + 6 = 0
II. 2y^{2} + 11y + 12 = 0
Correct
6x^{2} + 13x + 6 = 0
6x^{2} + 9x + 4x + 6 = 0
Gives x = -2/3, -3/2
2y^{2} + 11y + 12 = 0
2y^{2} + 8y + 3y + 12 = 0
Gives y = -4, -3/2
Put on number line
-4… -3/2… -2/3
Incorrect
6x^{2} + 13x + 6 = 0
6x^{2} + 9x + 4x + 6 = 0
Gives x = -2/3, -3/2
2y^{2} + 11y + 12 = 0
2y^{2} + 8y + 3y + 12 = 0
Gives y = -4, -3/2
Put on number line
-4… -3/2… -2/3
Unattempted
6x^{2} + 13x + 6 = 0
6x^{2} + 9x + 4x + 6 = 0
Gives x = -2/3, -3/2
2y^{2} + 11y + 12 = 0
2y^{2} + 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. 6x^{2} + 5x – 1 = 0
II. 3y^{2} – 11y + 6 = 0
Correct
6x^{2} + 5x – 1 = 0
6x^{2} + 6x – x – 1 = 0
Gives x = -1, 1/6
3y^{2} – 11y + 6 = 0
3y^{2} – 9y – 2y + 6 = 0
Gives y = 2/3, 3
Put on number line
-1… 1/6… 2/3… 3
Incorrect
6x^{2} + 5x – 1 = 0
6x^{2} + 6x – x – 1 = 0
Gives x = -1, 1/6
3y^{2} – 11y + 6 = 0
3y^{2} – 9y – 2y + 6 = 0
Gives y = 2/3, 3
Put on number line
-1… 1/6… 2/3… 3
Unattempted
6x^{2} + 5x – 1 = 0
6x^{2} + 6x – x – 1 = 0
Gives x = -1, 1/6
3y^{2} – 11y + 6 = 0
3y^{2} – 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. 3x^{2} + 14x – 5 = 0
II. 3y^{2} – 19y + 6 = 0
Correct
3x^{2} + 14x – 5 = 0
3x^{2} + 15x – x – 5 = 0
Gives x = -5, 1/3
3y^{2} – 19y + 6 = 0
3y^{2} – 18y – y + 6 = 0
Gives y = 1/3, 6
Put on number line
-5…. 1/3… 6
Incorrect
3x^{2} + 14x – 5 = 0
3x^{2} + 15x – x – 5 = 0
Gives x = -5, 1/3
3y^{2} – 19y + 6 = 0
3y^{2} – 18y – y + 6 = 0
Gives y = 1/3, 6
Put on number line
-5…. 1/3… 6
Unattempted
3x^{2} + 14x – 5 = 0
3x^{2} + 15x – x – 5 = 0
Gives x = -5, 1/3
3y^{2} – 19y + 6 = 0
3y^{2} – 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. 4x^{2} + 4x – 15 = 0
II. y^{2} – 4y – 32 = 0
Correct
4x^{2} + 4x – 15 = 0
4x^{2} + 10x – 6x – 15 = 0
Gives x = -5/2, 3/2
y^{2} – 4y – 32 = 0
y^{2} + 4y – 8y – 32 = 0
Gives y= -4, 8
Put on number line
-4…. -5/2… 3/2… 8
Incorrect
4x^{2} + 4x – 15 = 0
4x^{2} + 10x – 6x – 15 = 0
Gives x = -5/2, 3/2
y^{2} – 4y – 32 = 0
y^{2} + 4y – 8y – 32 = 0
Gives y= -4, 8
Put on number line
-4…. -5/2… 3/2… 8
Unattempted
4x^{2} + 4x – 15 = 0
4x^{2} + 10x – 6x – 15 = 0
Gives x = -5/2, 3/2
y^{2} – 4y – 32 = 0
y^{2} + 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?
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?
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?
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?
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?
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?
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?
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
Correct
For 2 years, diff in CI and SI = Pr^{2}/100^{2}
For 3 years diff is Pr^{2}(r+300)/100^{3}
Pr^{2}/100^{2} / Pr^{2}(r+300)/100^{3}
= 15/47
So 100/(r+300) = 16/51
Solve, r = 300/16 = 18 3/4%
Incorrect
For 2 years, diff in CI and SI = Pr^{2}/100^{2}
For 3 years diff is Pr^{2}(r+300)/100^{3}
Pr^{2}/100^{2} / Pr^{2}(r+300)/100^{3}
= 15/47
So 100/(r+300) = 16/51
Solve, r = 300/16 = 18 3/4%
Unattempted
For 2 years, diff in CI and SI = Pr^{2}/100^{2}
For 3 years diff is Pr^{2}(r+300)/100^{3}
Pr^{2}/100^{2} / Pr^{2}(r+300)/100^{3}
= 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?
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.
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
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
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%
please provide detailed solution of approximation question
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
16,18,28 solns pls check
Yes
corrected
ty
@Shubhra_AspirantsZone:disqus q-15 , in q-17 E in place of D Mam