Directions(1 - 5): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.

A. if x > y B. if x ≤ y C. if x ≥ y D. if x < y E. if x = y or relationship between x and y can't be established

1. I. 5x[latex]^{2}[/latex] + 11x – 12 = 0, II. 4y[latex]^{2}[/latex] – 13y – 12 = 0
Answer: Option E
Explanation: I. 5x[latex]^{2}[/latex] + 11x – 12 = 0
⇒ 5x[latex]^{2}[/latex] + 15x – 4x – 12 = 0
⇒ 5x (x + 3) – 4(x + 3) = 0
⇒ (5x – 4) (x + 3) = 0
⇒ x = [latex]\frac{4}{5}[/latex], – 3
II. 4y[latex]^{2}[/latex] – 13y – 12 = 0
⇒ 4y[latex]^{2}[/latex] – 16y + 3y – 12 = 0
⇒ 4y(y – 4) + 3 (y – 4) = 0
⇒ (4y + 3) (y – 4) = 0
⇒ y = [latex]\frac{-3}{4}[/latex], 4
2. I. 3x[latex]^{2}[/latex] + 19x + 30 = 0, II. 3y[latex]^{2}[/latex] – 20y – 32 = 0
Answer: Option D
Explanation: I. 3x[latex]^{2}[/latex] + 19x + 30 = 0
⇒ 3x[latex]^{2}[/latex] + 9x + 10x + 30 = 0
⇒ 3x2 + 9x + 10x + 30 = 0
⇒ 3x (x + 3) + 10 (x + 3) = 0
⇒ (3x + 10)(x + 3) = 0
⇒ x = [latex]\frac{-10}{3}[/latex], – 3
II. 3y[latex]^{2}[/latex] – 20y – 32 = 0
⇒ 3y[latex]^{2}[/latex] – 24y + 4y – 32 = 0
⇒ 3y(y – 8) + 4(y – 8) = 0
⇒ (3y + 4) (y – 8) = 0
⇒ y = [latex]\frac{-4}{3}[/latex], 8
3. I. x[latex]^{2}[/latex] – 4√7x + 21 = 0, II. 2y[latex]^{2}[/latex] – 8√5y – 50 = 0
Answer: Option E
Explanation: I. x[latex]^{2}[/latex] – 4√7x + 21 = 0
⇒ x[latex]^{2}[/latex] – √7x – 3√7x + 21 = 0
⇒ x (x – √7) – 3√7 (x – √7) = 0
⇒ (x – √7)(x – 3√7) = 0
⇒ x = √7, 3√7
II. 2y[latex]^{2}[/latex] – 8√5y – 50 = 0
⇒ 2y[latex]^{2}[/latex] – 8√5y – 50 = 0
Taking 2 as a common term, we get
⇒ y[latex]^{2}[/latex] – 4√5y – 25 = 0
⇒ y[latex]^{2}[/latex] + √5y – 5√5y – 25 = 0
⇒ y( y + √5) – 5√5 (y + √5) = 0
⇒ (y + √5) (y – 5√5) = 0
⇒ y = – √5, 5√5
4. I. x[latex]^{2}[/latex]– 52x + 667 = 0, II. y[latex]^{2}[/latex] + 8y – 33 = 0
Answer: Option A
Explanation: x[latex]^{2}[/latex] – 52x + 667 = 0
⇒ x[latex]^{2}[/latex] – 23x – 29x + 667 = 0
⇒ x (x – 23) – 29 (x – 23) = 0
⇒ (x – 23) (x – 29) = 0
⇒ x = 23, 29
y[latex]^{2}[/latex] + 8y – 33 = 0
⇒ y[latex]^{2}[/latex] – 3y + 11y – 33 = 0
⇒ y (y – 3) + 11 (y – 3) = 0
⇒ (y – 3) (y + 11) = 0
⇒ y = 3, – 11
5. I. x[latex]^{2}[/latex] – 13 √2 x + 60 = 0, II. y[latex]^{2}[/latex] + 3√5 y – 20 = 0
Answer: Option A
Explanation: I. x[latex]^{2}[/latex] – 13√2 x + 60 = 0
⇒ x[latex]^{2}[/latex] – 10√2x – 3√2 x + 60 = 0
⇒ x(x – 10√2) – 3√2 (x – 10√2) = 0
⇒ (x – 3√2) (x – 10√2) = 0
x = 3√2, 10√2
II. y[latex]^{2}[/latex] + 3√5 y – 20 = 0
⇒ y[latex]^{2}[/latex] + 4√5 y – √5 y – 20 = 0
⇒ y(y + 4√5) – √5 (y + 4√5) = 0
⇒ (y – √5)(y + 4√5) = 0
⇒ y = – 4√5,√5

Directions(1 - 5): Each of the questions below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read both the statements and give answer:

A. The data in statements I alone is sufficient to answer the question, while the data in statement II alone is not sufficient to answer the question. B. The data in statements II alone is sufficient to answer the question, while the data in statement I alone is not sufficient to answer the question. C. The data in statements I alone or in statement II alone is sufficient to answer the question. D. The data in both the statements I and II is not sufficient to answer the question. E. The data in both the statements I and II together is necessary to answer the question.

1. What is the marked price of the shirt? I. A shopkeeper purchases 2 shirts for 2200 Rs. and earns 55 Rs. per shirt after giving 23% discount. II. A shopkeeper marks the price 36% more than the cost price and earns 200 Rs. after giving some discount.

A. 6912 B. 3456 C. 216 D. 69122 E. None of these

Answer: Option A
Explanation: From Statement I: Cost price of a shirt = 1100 Rs.
Selling price = 1100 + 55 = 1155 Rs.
Marked price = [latex]\frac{1155}{77}[/latex] × 100 = 1500 Rs.
Statement I is sufficient to answer the question.
From Statement II: Let the cost price = x Rs.
Marked price = x × 136% = 1.36x
Selling price = x + 200
Statement II is not sufficient to answer the question.
2. Find the height of the cylinder. I. The Curved surface area of the cylinder is 396 cm2 and the total surface area of the cylinder is 1628 cm2. II. The radius of the cylinder is 0.5 cm more than the 3 times of the height of the cylinder.

A. 20718 B. 18121 C. 16549 D. 14226 E. None of these

Answer: Option A
Explanation: From Statement I: Let radius = r, height = h
Curved surface area = 2πrh
396 = 2 × [latex]\frac{22}{7}[/latex] × r × h
r × h = 63
Total Surface area = 2πrh + 2πr[latex]^{2}[/latex]
1628 = [latex]\frac{22}{7}[/latex] (rh + r[latex]^{2}[/latex])
rh + r[latex]^{2}[/latex] = 259
63 + r[latex]^{2}[/latex] = 259
r[latex]^{2}[/latex] = 259 - 63 = 196
r = 14
r × h = 63
14 × h = 63
h = 4.5 cm
Statement I is alone sufficient to answer the question.
From Statement II:
r = 3h + 0.5
Statement II is not alone sufficient to answer the question.
3. Find the rate of interest. I. A sum of money becomes double of itself at a simple interest in 12.5 years. II. 5000 becomes 5832 in 2 years at compound interest. Answer: Option C
Explanation: From Statement I: Let sum = x ?, rate = y%
x = x × 12.5 × y%
y = 8%
Statement I is alone sufficient to answer the question.
From Statement II: A = P (1 + [latex]\frac{r}{100})^{t}[/latex]
5832 = (5000 + [latex]\frac{r}{100})^{2}[/latex]
[latex]\frac{5832}{5000}[/latex] = (1 + [latex]\frac{r}{100})^{2}[/latex]
[latex]\frac{729}{625}[/latex] = (1 + [latex]\frac{r}{100})^{2}[/latex]
[latex]\frac{27}{25}[/latex] = 1 + [latex]\frac{r}{100}[/latex]
[latex]\frac{27}{25}[/latex] – 1 = [latex]\frac{r}{100}[/latex]
[latex]\frac{2}{25}[/latex] = [latex]\frac{r}{100}[/latex]
r = 8%
Statement II is also alone sufficient to answer the question.
4. The total amount of Rs 62500 was distributed among two boys and 1 girl. How much did each boy get? I. Each girl got [latex]\frac{2}{3}[/latex]of what the two boys together got. II. The difference between the amount of two boys is Rs 12500. Answer: Option E
Explanation: Statement I: Let the boys got = Rs.a.
Girl got = a × [latex]\frac{2}{3}[/latex] = [latex]\frac{2a}{3}[/latex]
Statement II: Let boys got = x, x + 12500
Statement I + Statement II: x + x + 12500 + (x + x + 12500) × [latex]\frac{2}{3}[/latex] = 62500
2x + 12500 + (2x + 12500) × [latex]\frac{2}{3}[/latex] = 62500
2x + 12500 + [latex]\frac{4x}{3}[/latex] + [latex]\frac{2500}{3}[/latex] = 62500
6x + 37500 + 4x + 25000 = 62500 × 3
10x + 62500 = 187500
10x = 125000
x = 12500
Boys get = Rs.12500 , Rs.25000
Both the statements are necessary to give the answer.
5. What is the cost price of a suit? I. The marked price of 21 suits is Rs 10500. II. The profit earned by the shopkeeper on a suit is 20% more than the profit earned by the shopkeeper on a saree. Answer: Option D
Explanation: Statement I: Marked price of a suit = [latex]\frac{10500}{21}[/latex] = Rs.500
Statement II: Let the profit on a saree = x Rs
the profit on a suit = x × 120%
Statment I and Statement II both are not sufficient to answer the question.

Directions(1 - 5): Study the table to answer the given questions: Number of cars sold (in thousand) by 5 showrooms during 5 months 1. What is the average number of cars sold (in thousand) by showroom C in all the given months together?

A. 207 B. 211 C. 213 D.205 E. None of these

Answer: Option A
Explanation: Reqd avg = [latex]\frac{156 + 179 + 211 + 259 + 230}{5}[/latex]
= [latex]\frac{1035}{5}[/latex]
= 207 thousand
2. By what percent the number of cars sold by showroom B increase from March to July?

A. 24.33% B. 28.33% C. 26.33% D. 32.66% E.None of these

Answer: Option B
Explanation: Reqd % = [latex]\frac{231 – 180}{180}[/latex] × 100
= [latex]\frac{51}{180}[/latex] × 100
= 28.33%
3. The number of cars sold by showroom E increased by 8% from July to August and by 20% from August to September. How many cars were sold (in thousand) by store E in September?

A. 312 B. 318 C. 324 D. 328 E. None of these

Answer: Option C
Explanation: No. of cars sold by showroom E in July = 250
Now, as two percentage transactions are incurring on one specific base value we can apply the net % effect formula to get to know the overall % increase.
Net % effect = x + y + [latex]\frac{(xy)}{100}[/latex]
Putting the values we get,
Net % effect = 8 + 20 + [latex]\frac{(8 x 20)}{100}[/latex] = 29.6%
Therefore, the number of cars sold by E in September = 129.6% of 250 = 324
4. The total number of cars sold by all the given stores together in August was 7/9 of the total number of cars sold by all the given stores together in June. How many cars were sold (in thousand) by all the given stores together in August itself

A. 756 B. 798 C. 784 D. 826 E. None of these

Answer: Option C
Explanation: The number of cars sold by all stores in August = [latex]\frac{7}{9}[/latex] of (Cars sold by all stores in June)
= [latex]\frac{7}{9}[/latex] × (205 + 198 + 259 + 165 + 181)
= [latex]\frac{7}{9}[/latex] × 1008 = 7 × 112 = 784
5. The no. of cars sold by showroom D in July is approximately what percent more than the number of cars sold by show room A in March

A. 15.2% B. 16.4% C. 15.9% D. 15.6% E. None of these

Answer: Option D
Explanation: Reqd % = [latex]\frac{178 – 154}{154}[/latex] × 100
= [latex]\frac{24}{154}[/latex] × 100
= 15.6%