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Directions (6 – 10): In the following questions, two equations I and II are given. You have to solve both the equations and give answer as,
If x > y
If x ≥ y
If x < y
If x ≤ y
If x = y or the relation cannot be established
3x^{2} + 4√3x + 4 = 0
5y^{2} = 2√5y + 8
Correct
Answer: c)
I. 3x^{2} + 4√3x + 4 = 0
3x^{2} + 2√3x + 2√3x + 4 = 0
√3x (√3x + 2) + 2(√3x + 2) = 0
(√3x + 2) (√3x + 2) = 0
x = -2/√3, -2/√3
II. 5y^{2} – 2√5y – 8 = 0
5y^{2} – 4√5y + 2√5y – 8 = 0
√5y (√5y – 4) + 2(√5y – 4) = 0
(√5y + 2) (√5y – 4) = 0
y = -2/√5, 4/√5
x < y
Incorrect
Answer: c)
I. 3x^{2} + 4√3x + 4 = 0
3x^{2} + 2√3x + 2√3x + 4 = 0
√3x (√3x + 2) + 2(√3x + 2) = 0
(√3x + 2) (√3x + 2) = 0
x = -2/√3, -2/√3
II. 5y^{2} – 2√5y – 8 = 0
5y^{2} – 4√5y + 2√5y – 8 = 0
√5y (√5y – 4) + 2(√5y – 4) = 0
(√5y + 2) (√5y – 4) = 0
y = -2/√5, 4/√5
x < y
Question 7 of 20
7. Question
Directions (6 – 10): In the following questions, two equations I and II are given. You have to solve both the equations and give answer as,
If x > y
If x ≥ y
If x < y
If x ≤ y
If x = y or the relation cannot be established
8x^{2} + 42x + 45 = 0
2y^{2} + 34y + 104 = 0
Correct
Answer: a)
I. 8x^{2} + 42x + 45 = 0
8x^{2} + 12x + 30x + 45 = 0
4x (2x + 3) + 15(2x + 3) = 0
(4x + 15) (2x + 3) = 0
x = -15/4, -3/2 = -3.75, -1.5
II. 2y^{2} + 34y + 104 = 0
2y^{2} + 8y + 26y + 104 = 0
2y(y + 4) + 26(y + 4) = 0
(2y + 26) (y + 4) = 0
y = -26/2, -4 = -13, -4
x > y
Incorrect
Answer: a)
I. 8x^{2} + 42x + 45 = 0
8x^{2} + 12x + 30x + 45 = 0
4x (2x + 3) + 15(2x + 3) = 0
(4x + 15) (2x + 3) = 0
x = -15/4, -3/2 = -3.75, -1.5
II. 2y^{2} + 34y + 104 = 0
2y^{2} + 8y + 26y + 104 = 0
2y(y + 4) + 26(y + 4) = 0
(2y + 26) (y + 4) = 0
y = -26/2, -4 = -13, -4
x > y
Question 8 of 20
8. Question
Directions (6 – 10): In the following questions, two equations I and II are given. You have to solve both the equations and give answer as,
If x > y
If x ≥ y
If x < y
If x ≤ y
If x = y or the relation cannot be established
3√x = (23/√x) + (13/√x)
(√y/4) + (4√y/8) = (9/√y)
Correct
Answer: e)
I. 3√x = (23/√x) + (13/√x)
3√x = (23 + 13)/ √x
x = 36/3 = 12
II. (√y/4) + (4√y/8) = (9/√y)
(2√y + 4√y)/8 = 9/√y
6√y /8 = 9/√y
y = (9/6)*8 = 12
x = y
Incorrect
Answer: e)
I. 3√x = (23/√x) + (13/√x)
3√x = (23 + 13)/ √x
x = 36/3 = 12
II. (√y/4) + (4√y/8) = (9/√y)
(2√y + 4√y)/8 = 9/√y
6√y /8 = 9/√y
y = (9/6)*8 = 12
x = y
Question 9 of 20
9. Question
Directions (6 – 10): In the following questions, two equations I and II are given. You have to solve both the equations and give answer as,
If x > y
If x ≥ y
If x < y
If x ≤ y
If x = y or the relation cannot be established
5x^{2} × x^{1/2} = 15625
y = ∛9261
Correct
Answer: a)
I. 5x^{2} × x^{1/2} = 15625
x^{2} × x^{1/2} = 15625/5
x^{2 + (½)} = 3125
x^{5/2} = 3125
x = (3125)^{2/5} = (5^{5})^{2/5}
x = 5^{2} = 25
II. y = ∛9261
Y = 21
X > y
Incorrect
Answer: a)
I. 5x^{2} × x^{1/2} = 15625
x^{2} × x^{1/2} = 15625/5
x^{2 + (½)} = 3125
x^{5/2} = 3125
x = (3125)^{2/5} = (5^{5})^{2/5}
x = 5^{2} = 25
II. y = ∛9261
Y = 21
X > y
Question 10 of 20
10. Question
Directions (6 – 10): In the following questions, two equations I and II are given. You have to solve both the equations and give answer as,
If x > y
If x ≥ y
If x < y
If x ≤ y
If x = y or the relation cannot be established
x^{2}+ 18x + 45 = 0
y^{2} + 19y + 60 = 0
Correct
Answer: e)
I. x^{2}+ 18x + 45 = 0
(x + 15) (x + 3) = 0
X = -15, -3
II. y^{2} + 19y + 60 = 0
(x + 15) (x + 4) = 0
X = -15, -4
Can’t be determined
Incorrect
Answer: e)
I. x^{2}+ 18x + 45 = 0
(x + 15) (x + 3) = 0
X = -15, -3
II. y^{2} + 19y + 60 = 0
(x + 15) (x + 4) = 0
X = -15, -4
Can’t be determined
Question 11 of 20
11. Question
A box contains 6 Pink balls, 3 black balls, 4 orange balls and 2 white balls. If four balls are drawn out randomly, then the probability of getting balls of different colour?
Correct
Answer: b)
Total probability n(S) = 15C_{4} = (15*14*13*12) / (1*2*3*4)
Required probability n(E) = 6C_{1} and 3C_{1} and 4C_{1} and 2C_{1}
P(E) = n(E)/n(S)
P(E) = (6C_{1} and 3C_{1} and 4C_{1 }and 2C_{1}) / 15C_{4}
P(E) = 48/455
Incorrect
Answer: b)
Total probability n(S) = 15C_{4} = (15*14*13*12) / (1*2*3*4)
Required probability n(E) = 6C_{1} and 3C_{1} and 4C_{1} and 2C_{1}
P(E) = n(E)/n(S)
P(E) = (6C_{1} and 3C_{1} and 4C_{1 }and 2C_{1}) / 15C_{4}
P(E) = 48/455
Question 12 of 20
12. Question
The ratio of 3 years after, the age of P and 5 years ago, the age of Q is 6: 5. The present age of R is 2 times the present age of P. The difference between the present age of Q and R is 10 years. Find the present age of P?
Correct
Answer: d)
The ratio of 3 years after, the age of P and 5 years ago, the age of Q = 6: 5 (6x, 5x)
Present age of P and Q = 6x – 3, 5x + 5
The age of R = 2*P = 2*(6x – 3) = 12x – 6
12x – 6 – (5x + 5) = 10
12x – 6 – 5x – 5 = 10
7x = 10 + 11
7x = 21
X = 3
Present age of P = 6x – 3 = 18 – 3 = 15 years
Incorrect
Answer: d)
The ratio of 3 years after, the age of P and 5 years ago, the age of Q = 6: 5 (6x, 5x)
Present age of P and Q = 6x – 3, 5x + 5
The age of R = 2*P = 2*(6x – 3) = 12x – 6
12x – 6 – (5x + 5) = 10
12x – 6 – 5x – 5 = 10
7x = 10 + 11
7x = 21
X = 3
Present age of P = 6x – 3 = 18 – 3 = 15 years
Question 13 of 20
13. Question
A, B and C can do a work in 18, 24 and 30 days respectively. They undertook to finish the work together for Rs. 11750. Find the difference between the shares of A and C?
Correct
Answer: a)
Ratio of Shares of A: B: C = (1/18): (1/24): (1/30) = 20: 15: 12
Difference between the shares of A and C = (8/47)*11750 = Rs. 2000
Incorrect
Answer: a)
Ratio of Shares of A: B: C = (1/18): (1/24): (1/30) = 20: 15: 12
Difference between the shares of A and C = (8/47)*11750 = Rs. 2000
Question 14 of 20
14. Question
A, B and C started a business by investing Rs. 20000, Rs. 35000 and Rs. 25000 respectively. After 4 months, A invested 15 % more than the initial investment and after 5 months, C invested Rs. 10000 more but B withdraw 20 % of the initial investment. Find the share of C, if the total profit at the end of the year is Rs. 82750?
The difference between the two numbers is 360. 28 % of one number is equal to 42 % of another number. Find the two numbers?
Correct
Answer: b)
Let the two numbers be x and y,
X – Y = 360
(28/100)*X = (42/100)*Y
(X/Y) = 3/2
1’s = 360
The two numbers be,
= > 3’s = (360*3) = 1080
= > 2’s = (360*2) = 720
Incorrect
Answer: b)
Let the two numbers be x and y,
X – Y = 360
(28/100)*X = (42/100)*Y
(X/Y) = 3/2
1’s = 360
The two numbers be,
= > 3’s = (360*3) = 1080
= > 2’s = (360*2) = 720
Question 16 of 20
16. Question
Directions (16 – 20) Study the following table carefully and answer the given questions:
Following table shows the total no of students appeared for the certain examination from different schools.
School
Total no of students appeared
Appeared boys : girls
Pass %
No of girls passed
A
2500
2 : 3
72 %
1100
B
3200
5 : 3
76 %
840
C
2800
3 : 4
85 %
1330
D
1600
5 : 3
68 %
513
E
1500
3 : 2
90 %
520
Total number of boys appeared for an examination from School A and B together is approximately what percentage of total number of girls appeared for an examination from School C and D together?
Correct
Answer: b)
Total number of boys appeared for an examination from School A and B together
= > 2500*(2/5) + 3200*(5/8)
= > 1000 + 2000 = 3000
Total number of girls appeared for an examination from School C and D together
= > 2800*(4/7) + 1600*(3/8)
= > 1600 + 600 = 2200
Required % = (3000/2200)*100 = 136 %
Incorrect
Answer: b)
Total number of boys appeared for an examination from School A and B together
= > 2500*(2/5) + 3200*(5/8)
= > 1000 + 2000 = 3000
Total number of girls appeared for an examination from School C and D together
= > 2800*(4/7) + 1600*(3/8)
= > 1600 + 600 = 2200
Required % = (3000/2200)*100 = 136 %
Question 17 of 20
17. Question
Directions (16 – 20) Study the following table carefully and answer the given questions:
Following table shows the total no of students appeared for the certain examination from different schools.
School
Total no of students appeared
Appeared boys : girls
Pass %
No of girls passed
A
2500
2 : 3
72 %
1100
B
3200
5 : 3
76 %
840
C
2800
3 : 4
85 %
1330
D
1600
5 : 3
68 %
513
E
1500
3 : 2
90 %
520
Find the ratio between the total number of boys passed from School A to that of School C?
Correct
Answer: d)
The total number of boys passed from School A
= > 2500*(72/100) – 1100
= > 1800 – 1100 = 700
The total number of boys passed from School C
= > 2800*(85/100) – 1330
= > 2380 – 1330 = 1050
Required ratio = 700: 1050 = 2: 3
Incorrect
Answer: d)
The total number of boys passed from School A
= > 2500*(72/100) – 1100
= > 1800 – 1100 = 700
The total number of boys passed from School C
= > 2800*(85/100) – 1330
= > 2380 – 1330 = 1050
Required ratio = 700: 1050 = 2: 3
Question 18 of 20
18. Question
Directions (16 – 20) Study the following table carefully and answer the given questions:
Following table shows the total no of students appeared for the certain examination from different schools.
School
Total no of students appeared
Appeared boys : girls
Pass %
No of girls passed
A
2500
2 : 3
72 %
1100
B
3200
5 : 3
76 %
840
C
2800
3 : 4
85 %
1330
D
1600
5 : 3
68 %
513
E
1500
3 : 2
90 %
520
Find the difference between the total number of students passed the examination from School C and E together to that of School B and D together?
Correct
Answer: a)
The total number of students passed the examination from School C and E together
= > 2800*(85/100) + 1500*(90/100)
= > 2380 + 1350 = 3730
The total number of students passed the examination from School B and D together
= > 3200*(76/100) + 1600*(68/100)
= > 2432 + 1088 = 3520
Required difference = 3730 – 3520 = 210
Incorrect
Answer: a)
The total number of students passed the examination from School C and E together
= > 2800*(85/100) + 1500*(90/100)
= > 2380 + 1350 = 3730
The total number of students passed the examination from School B and D together
= > 3200*(76/100) + 1600*(68/100)
= > 2432 + 1088 = 3520
Required difference = 3730 – 3520 = 210
Question 19 of 20
19. Question
Directions (16 – 20) Study the following table carefully and answer the given questions:
Following table shows the total no of students appeared for the certain examination from different schools.
School
Total no of students appeared
Appeared boys : girls
Pass %
No of girls passed
A
2500
2 : 3
72 %
1100
B
3200
5 : 3
76 %
840
C
2800
3 : 4
85 %
1330
D
1600
5 : 3
68 %
513
E
1500
3 : 2
90 %
520
Find the average number of boys passed the examination from School A, B, C and E together?
Correct
Answer: c)
The total number of boys passed the examination from School A, B, C and E together
Directions (16 – 20) Study the following table carefully and answer the given questions:
Following table shows the total no of students appeared for the certain examination from different schools.
School
Total no of students appeared
Appeared boys : girls
Pass %
No of girls passed
A
2500
2 : 3
72 %
1100
B
3200
5 : 3
76 %
840
C
2800
3 : 4
85 %
1330
D
1600
5 : 3
68 %
513
E
1500
3 : 2
90 %
520
Total number of girls passed from School A and B together is approximately what percentage more/less than the total number of girls passed from School C and E together?
Correct
Answer: e)
Total number of girls passed from School A and B together
= > 1100 + 840 = 1940
Total number of girls passed from School C and E together
= > 1330 + 520 = 1850
Required % = [(1940 – 1850)/1850]*100 = 5 % more
Incorrect
Answer: e)
Total number of girls passed from School A and B together
= > 1100 + 840 = 1940
Total number of girls passed from School C and E together
= > 1330 + 520 = 1850
Required % = [(1940 – 1850)/1850]*100 = 5 % more
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Directions (6 – 10): In the following questions, two equations I and II are given. You have to solve both the equations and give answer as,
If x > y
If x ≥ y
If x < y
If x ≤ y
If x = y or the relation cannot be established
6) I. 3x^{2} + 4√3x + 4 = 0
II. 5y^{2} = 2√5y + 8
7) I. 8x^{2} + 42x + 45 = 0
II. 2y^{2} + 34y + 104 = 0
8) I. 3√x = (23/√x) + (13/√x)
II. (√y/4) + (4√y/8) = (9/√y)
9) I. 5x^{2} × x^{1/2} = 15625
II. y = ∛9261
10) I. x^{2}+ 18x + 45 = 0
II. y^{2} + 19y + 60 = 0
11) A box contains 6 Pink balls, 3 black balls, 4 orange balls and 2 white balls. If four balls are drawn out randomly, then the probability of getting balls of different colour?
11/135
48/455
23/479
34/477
None of these
12) The ratio of 3 years after, the age of P and 5 years ago, the age of Q is 6: 5. The present age of R is 2 times the present age of P. The difference between the present age of Q and R is 10 years. Find the present age of P?
12 years
20 years
25 years
15 years
18 years
13) A, B and C can do a work in 18, 24 and 30 days respectively. They undertook to finish the work together for Rs. 11750. Find the difference between the shares of A and C?
Rs. 2000
Rs. 1800
Rs. 2400
Rs. 1200
None of these
14) A, B and C started a business by investing Rs. 20000, Rs. 35000 and Rs. 25000 respectively. After 4 months, A invested 15 % more than the initial investment and after 5 months, C invested Rs. 10000 more but B withdraw 20 % of the initial investment. Find the share of C, if the total profit at the end of the year is Rs. 82750?
Rs. 24600
Rs. 35400
Rs. 27500
Rs. 32800
None of these
15) The difference between the two numbers is 360. 28 % of one number is equal to 42 % of another number. Find the two numbers?
1160 and 800
1080 and 720
1230 and 870
950 and 590
None of these
Directions (16 – 20) Study the following table carefully and answer the given questions:
Following table shows the total no of students appeared for the certain examination from different schools.
School
Total no of students appeared
Appeared boys : girls
Pass %
No of girls passed
A
2500
2 : 3
72 %
1100
B
3200
5 : 3
76 %
840
C
2800
3 : 4
85 %
1330
D
1600
5 : 3
68 %
513
E
1500
3 : 2
90 %
520
16) Total number of boys appeared for an examination from School A and B together is approximately what percentage of total number of girls appeared for an examination from School C and D together?
115 %
136 %
100 %
85 %
150 %
17) Find the ratio between the total number of boys passed from School A to that of School C?
4: 5
5: 6
3: 4
2: 3
None of these
18) Find the difference between the total number of students passed the examination from School C and E together to that of School B and D together?
210
350
420
280
None of these
19) Find the average number of boys passed the examination from School A, B, C and E together?
2348
1579
1043
2156
None of these
20) Total number of girls passed from School A and B together is approximately what percentage more/less than the total number of girls passed from School C and E together?
10 % less
5 % less
15 % more
15 % less
5 % more
Answers:
Direction (1-5)
1). Answer: c)
47% of 1900 + 31% of 4500 = 27x + 1452 + 914 – 294