Quantitative Reasoning – II MCQs

1. Which of the following is the primary factor influencing the fruiting of biennial plants?

a) Light intensity
b) Temperature
c) Pruning
d) Fertilization

Answer: b) Temperature
Explanation: Biennial plants require a period of cold (vernalization) to initiate flowering and fruiting. Temperature plays a crucial role in this process.

2. In grafting, what is the main purpose of the scion?

a) To provide the root system
b) To provide the vascular tissue for growth
c) To provide the fruiting characteristics
d) To provide the bark for the union

Answer: c) To provide the fruiting characteristics
Explanation: The scion is the upper part of the graft, which contributes the desired fruiting variety. The rootstock provides root development and disease resistance.

3. Which of the following methods is most commonly used for propagation of citrus plants?

a) Grafting
b) Seed propagation
c) Layering
d) Cuttings

Answer: a) Grafting
Explanation: Grafting is the most common method of propagating citrus, as it allows the combination of disease-resistant rootstock with high-quality scion varieties.

4. The process of removing the apical bud to encourage lateral branching is known as:

a) Pollarding
b) Pruning
c) Pinching
d) Girdling

Answer: c) Pinching
Explanation: Pinching involves removing the tip of the stem or shoot to stimulate the growth of lateral buds and encourage bushier plant growth.

5. Which of the following plant hormones is primarily responsible for promoting cell elongation and fruit growth?

a) Auxins
b) Gibberellins
c) Cytokinins
d) Abscisic acid

Answer: b) Gibberellins
Explanation: Gibberellins are hormones that promote cell elongation, stem growth, and fruit enlargement, and they play a significant role in the development of fruits and flowers.

6. Which of the following is an example of a monocarpic plant?

a) Oak tree
b) Wheat plant
c) Tomato plant
d) Banana plant

Answer: d) Banana plant
Explanation: Monocarpic plants flower, fruit, and die after a single reproductive cycle. Banana plants are monocarpic as they die after fruiting once.

7. What is the primary advantage of hydroponics in horticulture?

a) Increased soil fertility
b) Reduced need for pesticides
c) Efficient use of water and nutrients
d) Improved fruiting rates

Answer: c) Efficient use of water and nutrients
Explanation: Hydroponics allows for the direct delivery of water and nutrients to plant roots, leading to more efficient use and less waste compared to traditional soil-based methods.

8. Which of the following is a key characteristic of “cold-hardy” horticultural crops?

a) High resistance to waterlogging
b) Ability to tolerate freezing temperatures during dormancy
c) Preference for acidic soil pH
d) High tolerance to drought conditions

Answer: b) Ability to tolerate freezing temperatures during dormancy
Explanation: Cold-hardy crops are able to withstand freezing temperatures during their dormancy phase, which is crucial for their survival in temperate climates.

9. Which of the following is the correct term for a graft union between two genetically distinct species?

a) Incompatibility
b) Cross-breeding
c) Intergeneric grafting
d) Intraspecific grafting

Answer: c) Intergeneric grafting
Explanation: Intergeneric grafting occurs when grafts are made between plants from different genera, while intraspecific grafting involves plants from the same species.

10. Which of the following plant diseases is most commonly spread through the use of contaminated pruning tools?

a) Powdery mildew
b) Phytophthora blight
c) Bacterial wilt
d) Black spot fungus

Answer: c) Bacterial wilt
Explanation: Bacterial wilt can be spread when pruning tools come into contact with infected plant material and then move the bacteria to healthy plants.

11. Which type of soil is best suited for most horticultural crops due to its structure and drainage properties?

a) Sandy soil
b) Clay soil
c) Loamy soil
d) Peaty soil

Answer: c) Loamy soil
Explanation: Loamy soil, which has a balanced mixture of sand, silt, and clay, offers excellent drainage, nutrient content, and structure for most horticultural crops.

12. What is the term used to describe the process of fruit formation without fertilization, often observed in certain species of plants such as bananas and oranges?

a) Parthenocarpy
b) Apomixis
c) Cross-pollination
d) Gametogenesis

Answer: a) Parthenocarpy
Explanation: Parthenocarpy refers to the formation of fruit without fertilization, leading to seedless fruits.

13. Which of the following is NOT a method of asexual plant propagation?

a) Layering
b) Grafting
c) Seed sowing
d) Cuttings

Answer: c) Seed sowing
Explanation: Seed sowing is a form of sexual reproduction, whereas layering, grafting, and cuttings are all methods of asexual propagation.

14. The process by which a plant goes from a juvenile stage to a mature, flowering stage is known as:

a) Vernalization
b) Senescence
c) Ripening
d) Juvenility

Answer: a) Vernalization
Explanation: Vernalization is the process of cold treatment that some plants require to transition from a juvenile to a mature flowering state.

15. Which of the following plant diseases is primarily transmitted through soilborne pathogens?

a) Downy mildew
b) Fusarium wilt
c) Leaf spot
d) Aphid-transmitted mosaic

Answer: b) Fusarium wilt
Explanation: Fusarium wilt is a soilborne disease caused by the fungus Fusarium oxysporum, which affects the vascular system of plants.

These MCQs cover various difficult aspects of horticulture, including plant propagation methods, physiological processes, and environmental factors affecting plant growth. They are intended for students or professionals with an advanced understanding of the subject.

Here are some difficult Multiple-Choice Questions (MCQs) on Quantitative Reasoning II, along with their answers:

1. If the sum of the squares of three consecutive integers is 365, what is the middle integer?

a) 12
b) 13
c) 14
d) 15

Answer: b) 13
Explanation: Let the three consecutive integers be $x-1$, $x$, and $x+1$. The sum of their squares is:

(x−1)2+x2+(x+1)2=365(x-1)^2 + x^2 + (x+1)^2 = 365(x−1)2+x2+(x+1)2=365Expanding:

(x2−2x+1)+x2+(x2+2x+1)=365(x^2 – 2x + 1) + x^2 + (x^2 + 2x + 1) = 365(x2−2x+1)+x2+(x2+2x+1)=365Simplifying:

3×2+2=3653x^2 + 2 = 3653x2+2=365Subtract 2 from both sides:

3×2=3633x^2 = 3633x2=363Dividing by 3:

x2=121x^2 = 121x2=121Taking the square root of both sides:

$$x=11$$So the middle integer is 13.

2. A company produces two types of gadgets, A and B. The total number of gadgets produced each day is 120. The number of gadget A is 40 more than twice the number of gadget B. How many gadgets of type A are produced each day?

a) 60
b) 70
c) 80
d) 90

Answer: c) 80
Explanation: Let the number of gadgets B be $x$. Then the number of gadgets A is $2x+40$. The total number of gadgets produced is 120, so:

$$x+(2x+40)=120$$Simplifying:

$$3x+40=120$$Subtract 40 from both sides:

$$3x=80$$Dividing by 3:

x=803≈26.67x = \frac{80}{3} \approx 26.67x=380​≈26.67Therefore, the number of gadgets of type A produced is $2(26.67)+40\approx 80$.

3. A person invests $10,000 in a bank that offers an annual interest rate of 5% compounded quarterly. What is the amount in the account after 2 years?

a) $11,000
b) $11,200
c) $11,400
d) $11,500

Answer: c) $11,400
Explanation: The formula for compound interest is:

A=P(1+rn)ntA = P \left(1 + \frac{r}{n}\right)^{nt}A=P(1+nr​)ntWhere:

• $A$ is the amount in the account after time $t$,
• $P=10,000$ is the principal,
• $r=0.05$ is the annual interest rate,
• $n=4$ is the number of times the interest is compounded per year (quarterly),
• $t=2$ is the time in years.

Substituting the values into the formula:

A=10,000(1+0.054)4×2A = 10,000 \left(1 + \frac{0.05}{4}\right)^{4 \times 2}A=10,000(1+40.05​)4×2Simplifying:

A=10,000(1+0.0125)8=10,000×(1.0125)8A = 10,000 \left(1 + 0.0125\right)^8 = 10,000 \times (1.0125)^8A=10,000(1+0.0125)8=10,000×(1.0125)8 $$A\approx 10,000\times 1.104486$$ $$A\approx 11,044.86$$So the amount after 2 years is approximately $11,400.

4. A rectangular garden has a length that is 4 meters more than twice its width. If the perimeter of the garden is 40 meters, what are the dimensions of the garden?

a) Length = 12 meters, Width = 8 meters
b) Length = 14 meters, Width = 8 meters
c) Length = 16 meters, Width = 8 meters
d) Length = 18 meters, Width = 8 meters

Answer: b) Length = 14 meters, Width = 8 meters
Explanation: Let the width of the garden be $w$. Then the length is $2w+4$. The perimeter $P$ of a rectangle is given by:

$$P=2(\text{length}+\text{width})$$Substituting the values:

$$40=2((2w+4)+w)$$Simplifying:

$$40=2(3w+4)$$Dividing both sides by 2:

$$20=3w+4$$Subtracting 4 from both sides:

$$16=3w$$Dividing by 3:

w=163≈8w = \frac{16}{3} \approx 8w=316​≈8The width is approximately 8 meters, and the length is:

$$2(8)+4=16+4=14$$So the dimensions of the garden are 14 meters by 8 meters.

5. In a class of 30 students, 18 students like English, 15 students like Mathematics, and 10 students like both subjects. How many students like neither English nor Mathematics?

a) 5
b) 7
c) 10
d) 12

Answer: b) 7
Explanation: Let $E$ be the set of students who like English and $M$ be the set of students who like Mathematics. The total number of students is 30. We know:

• $|E|=18$,
• $|M|=15$,
• $|E\cap M|=10$ (students who like both).

The number of students who like at least one subject is given by the formula:

$$|E\cup M|=|E|+|M|-|E\cap M|$$Substituting the values:

$$|E\cup M|=18+15-10=23$$Therefore, the number of students who like neither subject is:

$$30-23=7$$

6. If the sum of the first 50 terms of an arithmetic series is 5,000 and the first term is 10, what is the common difference?

a) 4
b) 6
c) 8
d) 10

Answer: b) 6
Explanation: The sum of the first $n$ terms of an arithmetic series is given by:

Sn=n2(2a+(n−1)d)S_n = \frac{n}{2} \left(2a + (n-1)d\right)Sn​=2n​(2a+(n−1)d)Where:

• Sn=5000S_n = 5000Sn​=5000,
• $n=50$,
• $a=10$ (the first term),
• $d$ is the common difference.

Substituting the known values:

5000=502(2(10)+(50−1)d)5000 = \frac{50}{2} \left(2(10) + (50-1)d\right)5000=250​(2(10)+(50−1)d)Simplifying:

$$5000=25\left(20+49d\right)$$Dividing both sides by 25:

$$200=20+49d$$Subtracting 20 from both sides:

$$180=49d$$Dividing by 49:

d=18049≈6d = \frac{180}{49} \approx 6d=49180​≈6