“20-20” Quantitative Aptitude | Crack Indian bank PO Prelims 2018 Day-182
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“20-20” Aptitude Questions | Crack Indian bank PO prelims 2018 Day-182
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Question 1 of 20
1. Question
Directions (1 – 5): In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,
a) If x > y
b) If x ≥ y
c) If x < y
d) If x ≤ y
e) If x = y or the relation cannot be established
x^{2} – 3x – 28 = 0
y^{2} – y – 72 = 0
Correct
Answer: e)
x^{2} – 3x – 28 = 0
x^{2} – 7x + 4x – 28 = 0
x(x – 7)+4(x – 7) = 0
(x + 4) (x – 7) = 0
X = -4, 7
II. y^{2} – y – 72 = 0
y^{2} – 9y + 8y – 72 = 0
y(y – 9)+ 8(y – 9) =0
(y + 8)(y – 9) = 0
Y = -8, 9
Can’t be determined
Incorrect
Answer: e)
x^{2} – 3x – 28 = 0
x^{2} – 7x + 4x – 28 = 0
x(x – 7)+4(x – 7) = 0
(x + 4) (x – 7) = 0
X = -4, 7
II. y^{2} – y – 72 = 0
y^{2} – 9y + 8y – 72 = 0
y(y – 9)+ 8(y – 9) =0
(y + 8)(y – 9) = 0
Y = -8, 9
Can’t be determined
Question 2 of 20
2. Question
Directions (1 – 5): In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,
a) If x > y
b) If x ≥ y
c) If x < y
d) If x ≤ y
e) If x = y or the relation cannot be established
3x^{2} + 19x + 28 = 0
2y^{2} + 13y + 21 = 0
Correct
Answer: e)
3x^{2} + 19x + 28 = 0
3x^{2} + 12x + 7x + 28 = 0
3x(x + 4) + 7(x + 4) = 0
(3x + 7) (x + 4) = 0
X = -7/3, -4 = -2.33, – 4
II. 2y^{2} + 13y + 21 = 0
2y^{2} + 6y + 7y + 21 = 0
2y(y + 3) +7(y + 3) =0
(2y + 7)(y + 3) = 0
Y = -7/2, -3 = -3.5, -3
Can’t be determined
Incorrect
Answer: e)
3x^{2} + 19x + 28 = 0
3x^{2} + 12x + 7x + 28 = 0
3x(x + 4) + 7(x + 4) = 0
(3x + 7) (x + 4) = 0
X = -7/3, -4 = -2.33, – 4
II. 2y^{2} + 13y + 21 = 0
2y^{2} + 6y + 7y + 21 = 0
2y(y + 3) +7(y + 3) =0
(2y + 7)(y + 3) = 0
Y = -7/2, -3 = -3.5, -3
Can’t be determined
Question 3 of 20
3. Question
Directions (1 – 5): In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,
a) If x > y
b) If x ≥ y
c) If x < y
d) If x ≤ y
e) If x = y or the relation cannot be established
2x-3y = -6
3x + 4y = 25
Correct
Answer: c)
2x-3y = -6-à(1)
3x + 4y = 25-à(2)
By solving the equation (1) and (2),
X = 3, y = 4
X < y
Incorrect
Answer: c)
2x-3y = -6-à(1)
3x + 4y = 25-à(2)
By solving the equation (1) and (2),
X = 3, y = 4
X < y
Question 4 of 20
4. Question
Directions (1 – 5): In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,
a) If x > y
b) If x ≥ y
c) If x < y
d) If x ≤ y
e) If x = y or the relation cannot be established
12x^{2}– 37x + 21 = 0
15y^{2} + 54y + 27 = 0
Correct
Answer: a)
12x^{2}– 37x + 21 = 0
12x^{2}– 28x – 9x + 21 = 0
4x (3x – 7) – 3(3x – 7) = 0
(4x – 3) (3x – 7) = 0
X = ¾, 7/3 = 0.75, 2.33
II. 15y^{2} + 54y + 27 = 0
15y^{2} + 45y+ 9y + 27 = 0
15y(y + 3) + 9(y + 3) = 0
(15y + 9) (y + 3) = 0
Y = -9/15, -3 = -3/5, -3
X > y
Incorrect
Answer: a)
12x^{2}– 37x + 21 = 0
12x^{2}– 28x – 9x + 21 = 0
4x (3x – 7) – 3(3x – 7) = 0
(4x – 3) (3x – 7) = 0
X = ¾, 7/3 = 0.75, 2.33
II. 15y^{2} + 54y + 27 = 0
15y^{2} + 45y+ 9y + 27 = 0
15y(y + 3) + 9(y + 3) = 0
(15y + 9) (y + 3) = 0
Y = -9/15, -3 = -3/5, -3
X > y
Question 5 of 20
5. Question
Directions (1 – 5): In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,
a) If x > y
b) If x ≥ y
c) If x < y
d) If x ≤ y
e) If x = y or the relation cannot be established
x^{2} + 3√7 x – 70 = 0
y^{2} + 2√3 y – 105 = 0
Correct
Answer: e)
x^{2} + 3√7 x – 70 = 0
x^{2} + 5√7 x – 2√7 x – 70 = 0
x (x + 5√7) – 2√7 (x + 5√7) = 0
(x – 2√7) (x + 5√7) = 0
X = 2√7, – 5√7
II. y^{2} + 2√3 y – 105 = 0
y^{2} + 7√3 y – 5√3 y – 105 = 0
y(y + 7√3) – 5√3 (y + 7√3) = 0
(y – 5√3) (y + 7√3) = 0
Y = 5√3, – 7√3
Can’t be determined
Incorrect
Answer: e)
x^{2} + 3√7 x – 70 = 0
x^{2} + 5√7 x – 2√7 x – 70 = 0
x (x + 5√7) – 2√7 (x + 5√7) = 0
(x – 2√7) (x + 5√7) = 0
X = 2√7, – 5√7
II. y^{2} + 2√3 y – 105 = 0
y^{2} + 7√3 y – 5√3 y – 105 = 0
y(y + 7√3) – 5√3 (y + 7√3) = 0
(y – 5√3) (y + 7√3) = 0
Y = 5√3, – 7√3
Can’t be determined
Question 6 of 20
6. Question
Directions (6 – 10): what value should come in place of question mark (?) in the following questions?
16 3/5 % of 17200 + 31 2/7 % of 2170 =?
Correct
Answer: c)
(83/500)*17200 + (219/700)*2170 = x
X = 2855.2 + 678.9
X = 3534.1
Incorrect
Answer: c)
(83/500)*17200 + (219/700)*2170 = x
X = 2855.2 + 678.9
X = 3534.1
Question 7 of 20
7. Question
Directions (6 – 10): what value should come in place of question mark (?) in the following questions?
96^{2} ÷ 8 × 3 + 5^{3} = 5 × ?
Correct
Answer: a)
(96*96*3)/8 + 125 = 5x
5x = 3456 + 125
5x = 3581
X = 716.2
Incorrect
Answer: a)
(96*96*3)/8 + 125 = 5x
5x = 3456 + 125
5x = 3581
X = 716.2
Question 8 of 20
8. Question
Directions (6 – 10): what value should come in place of question mark (?) in the following questions?
5 4/15 of 4350 + 6 6/19 of 3857 =? + 640
Correct
Answer: d)
(79/15)*4350 + (120/19)*3857 = x + 640
22910 + 24360 – 640 = x
X = 46630
Incorrect
Answer: d)
(79/15)*4350 + (120/19)*3857 = x + 640
22910 + 24360 – 640 = x
X = 46630
Question 9 of 20
9. Question
Directions (6 – 10): what value should come in place of question mark (?) in the following questions?
(5115 ÷ 31) + (4312 ÷ 11) + (3094 ÷ 14) = ?
Correct
Answer: b)
(5115/31) + (4312/11) + (3094/14) = x
X = 165 + 392 + 221
X = 778
Incorrect
Answer: b)
(5115/31) + (4312/11) + (3094/14) = x
X = 165 + 392 + 221
X = 778
Question 10 of 20
10. Question
Directions (6 – 10): what value should come in place of question mark (?) in the following questions?
(3/7) of (35/13) of 3055 + 63 = ?
Correct
Answer: e)
(3/7) * (35/13) * 3055 + 63 = x
3525 + 63 = x
X = 3525 + 63
X = 3588
Incorrect
Answer: e)
(3/7) * (35/13) * 3055 + 63 = x
3525 + 63 = x
X = 3525 + 63
X = 3588
Question 11 of 20
11. Question
The ratio of present age of Madhu and Deepa is 3 : x. Madhu is 9 years younger than Priya. Priya’s age after 11 years is 35 years. The sum of the present age of Madhu and Deepa is same as the age of Priya, after 16 years. What should come in place of X?
Correct
Answer: b)
The ratio of present age of Madhu and Deepa = 3 : x
Madhu = Priya – 9
Priya’s age after 11 years = 35 years
Priya’s present age = 24 years
Madhu’s present age = 24 – 9 = 15 years
3’s = 15
1’s = 5
According to the question,
15 + 5x = 40
5x = 25
X = 5
Incorrect
Answer: b)
The ratio of present age of Madhu and Deepa = 3 : x
Madhu = Priya – 9
Priya’s age after 11 years = 35 years
Priya’s present age = 24 years
Madhu’s present age = 24 – 9 = 15 years
3’s = 15
1’s = 5
According to the question,
15 + 5x = 40
5x = 25
X = 5
Question 12 of 20
12. Question
A sum of money invested for 7 years in Scheme 1 which offers SI at a rate of 8% per annum. The amount received from Scheme 1 after 7 years invested for 2 years in Scheme 2 which offers CI rate of 10% per annum. If the interest received from Scheme B was Rs.1638. What was the sum invested in Scheme 1?
Correct
Answer: b)
SI =>Amount = [(x*8*7)/100] + x = (56x+100x)/100 = 156x/100 = 39x/25
CI => (39x/25)*[(1+10/100)^{2} – 1] =1638
=> (39x/25)*[(121/100) – 1] = 1638
= > (39x/25)*[21/100] = 1638
= > X= (1638*100*25)/(21*39) = Rs. 5000
Incorrect
Answer: b)
SI =>Amount = [(x*8*7)/100] + x = (56x+100x)/100 = 156x/100 = 39x/25
CI => (39x/25)*[(1+10/100)^{2} – 1] =1638
=> (39x/25)*[(121/100) – 1] = 1638
= > (39x/25)*[21/100] = 1638
= > X= (1638*100*25)/(21*39) = Rs. 5000
Question 13 of 20
13. Question
Rajiv’s monthly income is 2 times the monthly income of Manoj. Monthly income of Manoj is 20 % more than the monthly income of Ragu. Monthly income of Ragu is Rs. 42000. Find the annual income of Rajiv?
Correct
Answer: d)
Rajiv’s monthly income = 2*Manoj’s monthly income
Manoj’s monthly income = (120/100)*Ragu’s monthly income
Monthly income of Ragu = Rs. 42000
Manoj’s monthly income = (120/100)*42000 = Rs. 50400
Rajiv’s monthly income = 2*50400 = Rs. 100800
Annual income of Rajiv = 100800*12 = Rs. 1209600
Incorrect
Answer: d)
Rajiv’s monthly income = 2*Manoj’s monthly income
Manoj’s monthly income = (120/100)*Ragu’s monthly income
Monthly income of Ragu = Rs. 42000
Manoj’s monthly income = (120/100)*42000 = Rs. 50400
Rajiv’s monthly income = 2*50400 = Rs. 100800
Annual income of Rajiv = 100800*12 = Rs. 1209600
Question 14 of 20
14. Question
A and B can complete a work in 9 days and 12 days respectively. A begins to do the work and they work alternately one at a time for one day each. The whole work will be completed by both of them in how many days?
Correct
Answer: a)
(A+B)’s two days work = 1/9 + 1/12 = 7/36
(A+B)’s ten days work = (7/36)*5 = 35/36
Remaining work = 1/36
A’s one day work = 1/9
1/36 work is done by A in = (1/36)*9 = 1/4 day
The whole work will be completed in 10 ¼ days
Incorrect
Answer: a)
(A+B)’s two days work = 1/9 + 1/12 = 7/36
(A+B)’s ten days work = (7/36)*5 = 35/36
Remaining work = 1/36
A’s one day work = 1/9
1/36 work is done by A in = (1/36)*9 = 1/4 day
The whole work will be completed in 10 ¼ days
Question 15 of 20
15. Question
The average marks of Maths of a class of 35 students is 54. If the marks of three students were misread as 45, 48 and 57 of the actual marks 55, 61 and 69 respectively, then what would be the correct average?
Correct
Answer: c)
The average marks of Maths of a class of 35 students = 54
Directions (16 – 20): Study the following information carefully and answer the given questions.
The following bar graph shows the percentage of growth in the population of two states.
Find the average percentage growth of population of Gujarat all the given years together?
Correct
Answer: c)
The average percentage growth in population of Gujarat all the given years together
= > (25 + 20 + 35 + 30 + 30 + 40)/6
= > 180/6 = 30
Incorrect
Answer: c)
The average percentage growth in population of Gujarat all the given years together
= > (25 + 20 + 35 + 30 + 30 + 40)/6
= > 180/6 = 30
Question 17 of 20
17. Question
Directions (16 – 20): Study the following information carefully and answer the given questions.
The following bar graph shows the percentage of growth in the population of two states.
If the population of Kerala in the year 2013 was 3. 6 lakhs, then find the population of Kerala in the year 2011?
Correct
Answer: a)
Let the population of Kerala in the year 2011 be x,
According to the question,
= > x*(125/100)*(120/100) = 3.6
= > x = (3.6*100*100)/(125*120)
= > x = 2.4 lakhs
Incorrect
Answer: a)
Let the population of Kerala in the year 2011 be x,
According to the question,
= > x*(125/100)*(120/100) = 3.6
= > x = (3.6*100*100)/(125*120)
= > x = 2.4 lakhs
Question 18 of 20
18. Question
Directions (16 – 20): Study the following information carefully and answer the given questions.
The following bar graph shows the percentage of growth in the population of two states.
If the population of Gujarat in the year 2014 was 5.6 lakhs and the population of Kerala in the year 2014 was 4 lakhs, then find the ratio of population of Gujarat to that of Kerala in the year 2016?
Correct
Answer: d)
The population of Gujarat in the year 2016
= > 5.6*(130/100)*(130/100)
= > 9.464 lakhs
The population of Kerala in the year 2016
= > 4*(135/100)*(130/100)
= > 7.02 lakhs
Required ratio = 9.464 : 7.02
= > 9464 : 7020 = 2366 : 1755
Incorrect
Answer: d)
The population of Gujarat in the year 2016
= > 5.6*(130/100)*(130/100)
= > 9.464 lakhs
The population of Kerala in the year 2016
= > 4*(135/100)*(130/100)
= > 7.02 lakhs
Required ratio = 9.464 : 7.02
= > 9464 : 7020 = 2366 : 1755
Question 19 of 20
19. Question
Directions (16 – 20): Study the following information carefully and answer the given questions.
The following bar graph shows the percentage of growth in the population of two states.
The percentage growth of population of Kerala in the year 2011 and 2012 together is approximately what percentage more than the percentage growth of population of Gujarat in the year 2013?
Correct
Answer: b)
The percentage growth of population of Kerala in the year 2011 and 2012 together
= > 15 + 25 = 40 %
The percentage growth of population of Gujarat in the year 2013 = 35 %
Required % = [(40 – 35)/35]*100 = 14.28 % = 15 %
Incorrect
Answer: b)
The percentage growth of population of Kerala in the year 2011 and 2012 together
= > 15 + 25 = 40 %
The percentage growth of population of Gujarat in the year 2013 = 35 %
Required % = [(40 – 35)/35]*100 = 14.28 % = 15 %
Question 20 of 20
20. Question
Directions (16 – 20): Study the following information carefully and answer the given questions.
The following bar graph shows the percentage of growth in the population of two states.
Find the difference between the total population of Gujarat in 2014 to that of total population of Kerala in 2015, if the total population of Gujarat in 2015 is 5.2 lakhs and the population of Kerala in 2016 is 6.5 lakhs?
Correct
Answer: a)
Let the population of Gujarat in 2014 and that of Kerala in 2015 be x and y respectively,
The total population of Gujarat in 2015 = 5.2 lakhs
X*(130/100) = 5.2
X = 5.2*(100/130) = 4 lakhs
The population of Kerala in 2016 = 6.5 lakhs
Y*(130/100) = 6.5
Y = 6.5*(100/130) = 5 lakhs
Required difference = 5 – 4 = 1 lakh
Incorrect
Answer: a)
Let the population of Gujarat in 2014 and that of Kerala in 2015 be x and y respectively,
The total population of Gujarat in 2015 = 5.2 lakhs
X*(130/100) = 5.2
X = 5.2*(100/130) = 4 lakhs
The population of Kerala in 2016 = 6.5 lakhs
Y*(130/100) = 6.5
Y = 6.5*(100/130) = 5 lakhs
Required difference = 5 – 4 = 1 lakh
Click “Start Quiz” to attend these Questions and view Solutions
Directions (1 – 5): In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,
a) If x > y
b) If x ≥ y
c) If x < y
d) If x ≤ y
e) If x = y or the relation cannot be established
1) I. x^{2} – 3x – 28 = 0
II. y^{2} – y – 72 = 0
2) I. 3x^{2} + 19x + 28 = 0
II. 2y^{2} + 13y + 21 = 0
3) I. 2x-3y = -6
II. 3x + 4y = 25
4) I. 12x^{2}– 37x + 21 = 0
II. 15y^{2} + 54y + 27 = 0
5) I. x^{2} + 3√7 x – 70 = 0
II. y^{2} + 2√3 y – 105 = 0
Directions (6 – 10): what value should come in place of question mark (?) in the following questions?
6) 16 3/5 % of 17200 + 31 2/7 % of 2170 =?
a) 2857.6
b) 3132.4
c) 3534.1
d) 1878.8
e) None of these
7) 96^{2} ÷ 8 × 3 + 5^{3} = 5 × ?
a) 716.2
b) 852.8
c) 934.5
d) 558.7
e) None of these
8) 5 4/15 of 4350 + 6 6/19 of 3857 =? + 640
a) 41520
b) 38750
c) 32670
d) 46630
e) None of these
9) (5115 ÷ 31) + (4312 ÷ 11) + (3094 ÷ 14) = ?
a) 852
b) 778
c) 664
d) 526
e) None of these
10) (3/7) of (35/13) of 3055 + 63 = ?
a) 4256
b) 4890
c) 3924
d) 2962
e) 3588
11) The ratio of present age of Madhu and Deepa is 3 : x. Madhu is 9 years younger than Priya. Priya’s age after 11 years is 35 years. The sum of the present age of Madhu and Deepa is same as the age of Priya, after 16 years. What should come in place of X?
a) 4
b) 5
c) 3
d) 6
e) None of these
12) A sum of money invested for 7 years in Scheme 1 which offers SI at a rate of 8% per annum. The amount received from Scheme 1 after 7 years invested for 2 years in Scheme 2 which offers CI rate of 10% per annum. If the interest received from Scheme B was Rs.1638. What was the sum invested in Scheme 1?
a) Rs.7500
b) Rs.5000
c) Rs.8200
d) Rs.9000
e) None of these
13) Rajiv’s monthly income is 2 times the monthly income of Manoj. Monthly income of Manoj is 20 % more than the monthly income of Ragu. Monthly income of Ragu is Rs. 42000. Find the annual income of Rajiv?
a) Rs. 1056800
b) Rs. 924600
c) Rs. 837500
d) Rs. 1209600
e) None of these
14) A and B can complete a work in 9 days and 12 days respectively. A begins to do the work and they work alternately one at a time for one day each. The whole work will be completed by both of them in how many days?
a) 10 ¼ days
b) 9 3/5 days
c) 12 5/7 days
d) 11 ½ days
e) None of these
15) The average marks of Maths of a class of 35 students is 54. If the marks of three students were misread as 45, 48 and 57 of the actual marks 55, 61 and 69 respectively, then what would be the correct average?
a) 56
b) 55.5
c) 55
d) 55.25
e) None of these
Directions (16 – 20): Study the following information carefully and answer the given questions.
The following bar graph shows the percentage of growth in the population of two states.
16) Find the average percentage growth of population of Gujarat all the given years together?
a) 28
b) 26
c) 30
d) 23
e) None of these
17) If the population of Kerala in the year 2013 was 3. 6 lakhs, then find the population of Kerala in the year 2011?
a) 2.4 lakhs
b) 2.5 lakhs
c) 2.8 lakhs
d) 3.2 lakhs
e) None of these
18) If the population of Gujarat in the year 2014 was 5.6 lakhs and the population of Kerala in the year 2014 was 4 lakhs, then find the ratio of population of Gujarat to that of Kerala in the year 2016?
a) 1352 : 1089
b) 567: 352
c) 15: 8
d) 2366: 1755
e) None of these
19) The percentage growth of population of Kerala in the year 2011 and 2012 together is approximately what percentage more than the percentage growth of population of Gujarat in the year 2013?
a) 20 %
b) 15 %
c) 10 %
d) 25 %
e) 30 %
20) Find the difference between the total population of Gujarat in 2014 to that of total population of Kerala in 2015, if the total population of Gujarat in 2015 is 5.2 lakhs and the population of Kerala in 2016 is 6.5 lakhs?
a) 1 lakh
b) 1.5 lakh
c) 2 lakhs
d) 2.25 lakhs
e) None of these
Answers :
Direction (1-5) :
1). Answer: e)
x^{2} – 3x – 28 = 0
x^{2} – 7x + 4x – 28 = 0
x(x – 7)+4(x – 7) = 0
(x + 4) (x – 7) = 0
X = -4, 7
II. y^{2} – y – 72 = 0
y^{2} – 9y + 8y – 72 = 0
y(y – 9)+ 8(y – 9) =0
(y + 8)(y – 9) = 0
Y = -8, 9
Can’t be determined
2). Answer: e)
3x^{2} + 19x + 28 = 0
3x^{2} + 12x + 7x + 28 = 0
3x(x + 4) + 7(x + 4) = 0
(3x + 7) (x + 4) = 0
X = -7/3, -4 = -2.33, – 4
II. 2y^{2} + 13y + 21 = 0
2y^{2} + 6y + 7y + 21 = 0
2y(y + 3) +7(y + 3) =0
(2y + 7)(y + 3) = 0
Y = -7/2, -3 = -3.5, -3
Can’t be determined
3). Answer: c)
2x-3y = -6-à(1)
3x + 4y = 25-à(2)
By solving the equation (1) and (2),
X = 3, y = 4
X < y
4). Answer: a)
12x^{2}– 37x + 21 = 0
12x^{2}– 28x – 9x + 21 = 0
4x (3x – 7) – 3(3x – 7) = 0
(4x – 3) (3x – 7) = 0
X = ¾, 7/3 = 0.75, 2.33
II. 15y^{2} + 54y + 27 = 0
15y^{2} + 45y+ 9y + 27 = 0
15y(y + 3) + 9(y + 3) = 0
(15y + 9) (y + 3) = 0
Y = -9/15, -3 = -3/5, -3
X > y
5). Answer: e)
x^{2} + 3√7 x – 70 = 0
x^{2} + 5√7 x – 2√7 x – 70 = 0
x (x + 5√7) – 2√7 (x + 5√7) = 0
(x – 2√7) (x + 5√7) = 0
X = 2√7, – 5√7
II. y^{2} + 2√3 y – 105 = 0
y^{2} + 7√3 y – 5√3 y – 105 = 0
y(y + 7√3) – 5√3 (y + 7√3) = 0
(y – 5√3) (y + 7√3) = 0
Y = 5√3, – 7√3
Can’t be determined
Direction (6-10) :
6). Answer: c)
(83/500)*17200 + (219/700)*2170 = x
X = 2855.2 + 678.9
X = 3534.1
7). Answer: a)
(96*96*3)/8 + 125 = 5x
5x = 3456 + 125
5x = 3581
X = 716.2
8). Answer: d)
(79/15)*4350 + (120/19)*3857 = x + 640
22910 + 24360 – 640 = x
X = 46630
9). Answer: b)
(5115/31) + (4312/11) + (3094/14) = x
X = 165 + 392 + 221
X = 778
10). Answer: e)
(3/7) * (35/13) * 3055 + 63 = x
3525 + 63 = x
X = 3525 + 63
X = 3588
11). Answer: b)
The ratio of present age of Madhu and Deepa = 3 : x
Madhu = Priya – 9
Priya’s age after 11 years = 35 years
Priya’s present age = 24 years
Madhu’s present age = 24 – 9 = 15 years
3’s = 15
1’s = 5
According to the question,
15 + 5x = 40
5x = 25
X = 5
12). Answer: b)
SI =>Amount = [(x*8*7)/100] + x = (56x+100x)/100 = 156x/100 = 39x/25
CI => (39x/25)*[(1+10/100)^{2} – 1] =1638
=> (39x/25)*[(121/100) – 1] = 1638
= > (39x/25)*[21/100] = 1638
= > X= (1638*100*25)/(21*39) = Rs. 5000
13). Answer: d)
Rajiv’s monthly income = 2*Manoj’s monthly income
Manoj’s monthly income = (120/100)*Ragu’s monthly income
Monthly income of Ragu = Rs. 42000
Manoj’s monthly income = (120/100)*42000 = Rs. 50400
Rajiv’s monthly income = 2*50400 = Rs. 100800
Annual income of Rajiv = 100800*12 = Rs. 1209600
14). Answer: a)
(A+B)’s two days work = 1/9 + 1/12 = 7/36
(A+B)’s ten days work = (7/36)*5 = 35/36
Remaining work = 1/36
A’s one day work = 1/9
1/36 work is done by A in = (1/36)*9 = 1/4 day
The whole work will be completed in 10 ¼ days
15). Answer: c)
The average marks of Maths of a class of 35 students = 54