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Question 151

KVS PRT 2023• Subject Specific

A boy goes to School at a speed of 6 km/hr which is 4 km away from home and returns home at a speed of 12 km/hr. The average speed is :

  1. AOption (a)

    9 km/hr

  2. BOption (b)

    8 km/hr

  3. COption (c)

    6 km/hr

  4. DOption (d)

    10 km/hr

View answer & explanation

Correct answer

Answer: (b)B.

8 km/hr

Why this answer is correct

.Given :

Distance from home to school = 4 km

Speed from home to school = 6 km/hr

Speed from school to home = 12 km/hr

Time taken = Distance / Speed Time taken = 4 km / 6 km/hr Time taken = 2/3 hours = 0.667 hours

Time taken = Distance / Speed Time taken = 4 km / 12 km/hr Time taken = 1/3 hours = 0.333 hours

Total distance = Distance from home to school + Distance from school to home

Total distance = 4 km + 4 km = 8 km

Total time taken = Time taken from home to school + Time taken from school to home

Total time taken = 0.667 hours + 0.333 hours = 1 hour

Average speed = Total distance / Total time taken Average speed = 8 km / 1 hour Average speed = 8 km/hr

Therefore, the average speed is 8 km/hr.

Hence, the correct answer is option 2) 8 km/hr.

Question 152

KVS PRT 2023• Subject Specific

A bunch of 8 bananas costs Rs. 24 and a single banana costs Rs. 4.50. Raju purchased 66 bananas. The amount of money spent by him is :

  1. AOption (a)

    Rs. 297

  2. BOption (b)

    Rs. 291

  3. COption (c)

    Rs. 201

  4. DOption (d)

    Rs. 198

View answer & explanation

Correct answer

Answer: (c)C.

Rs. 201

Why this answer is correct

Calculations:

To determine the amount of money Raju spent on purchasing 66 bananas, we need to consider the cost of bananas purchased in bunches and as single bananas.

First, let's calculate the cost per banana when bought in a bunch:

Cost per banana in a bunch = 24/8 = 3 

Given that the cost of a single banana is Rs. 4.50, it's clear that buying bananas in bunches is cheaper.

So, Raju should buy as many bananas in bunches as possible.

We will calculate the number of bunches Raju can buy:

Number of bunches = 66/8 with a remainder of 2 bananas

The cost for 8 bunches:

Cost for 8 bunches = 8 ×24 = 192 Rs

 

The remaining 2 bananas will be bought as single bananas:

Cost for 2 single banana s = 2 × 4.50 = 9

Total amount spent by Raju:

Total cost = 192 + 9  = 201 Rs

Question 153

KVS PRT 2023• Subject Specific

8.4 litre of a medicine is to be packed in small bottles of capacity 75 ml. The number of small bottles which can be filled is:

  1. AOption (a)

    100

  2. BOption (b)

    110

  3. COption (c)

    112

  4. DOption (d)

    120

View answer & explanation

Correct answer

Answer: (c)C.

112

Why this answer is correct

Concept use:

1 litre = 1000 millilitres

Explanation:

To determine the number of small bottles needed to pack 8.4 liters of medicine, with each bottle having a capacity of 75 milliliters, we first need to convert the volume from liters to milliliters. 

Since 1 liter is equivalent to 1000 millilitres:

8.4 litres= 8.4 ×  1000  millilitres = 8400 millilitres

Next, we divide the total volume in milliliters by the capacity of each bottle:

Number of bottles = 8400/75 = 112

Thus, the number of small bottles required to pack 8.4 liters of medicine is: 112.

Question 154

KVS PRT 2023• Subject Specific

In a school, 44% of the students are girls and the rest are boys. If the number of boys is 448, find the number of girls in the school.

  1. AOption (a)

    800

  2. BOption (b)

    400

  3. COption (c)

    360

  4. DOption (d)

    352

View answer & explanation

Correct answer

Answer: (d)D.

352

Why this answer is correct

Given :

44% of the students are girls.

The number of boys is 448.

Let the total number of students be x.

Number of boys = 448 Number of girls = x - 448 (since the rest are girls)

We know that the percentage of girls is 44%.

44% of x = x - 448 0.44x = x - 448 0.56x = 448 x = 448 / 0.56 x = 800

 Number of girls = 44% of the total number of students Number of girls = 0.44 × 800 Number of girls = 352

Hence, the number of girls in the school is 352.

Question 155

KVS PRT 2023• Subject Specific

Sujata bought \(\rm \frac{2}{3} \ kg\) of brinjals, \(\rm 2\frac{1}{3} \ kg\) of potatoes, \(\rm 2\frac{1}{2} \ kg\) of onions and 600 g of ginger. The total weight of vegetables bought by her (in decimals) is:

  1. AOption (a)

    5.9 kg

  2. BOption (b)

    6.0 kg

  3. COption (c)

    6.1 kg

  4. DOption (d)

    6.25 kg

View answer & explanation

Correct answer

Answer: (c)C.

6.1 kg

Why this answer is correct

Calculation:

Step 1: Convert mixed numbers to decimals:

\(\frac{2}{3} \text{ kg} = 0.67 \text{ kg} \)

\(2\frac{1}{3} \text{ kg} = 2 + \frac{1}{3} = 2 + 0.33 = 2.33 \text{ kg} \)

\(2\frac{1}{2} \text{ kg} = 2 + \frac{1}{2} = 2 + 0.50 = 2.50 \text{ kg} \)

600 g = \(\frac{600}{1000} \text{ kg} = 0.60 \text{ kg}\)

Sum the weights:

Total weight = \(0.67 \text{ kg} + 2.33 \text{ kg} + 2.50 \text{ kg} + 0.60 \text{ kg}\)

Total weight = 0.67 + 2.33 + 2.50 + 0.60 = 6.10 

So, the total weight of vegetables bought by Sujata is 6.10 kg.

Question 156

KVS PRT 2023• Subject Specific

\(\frac{1}{2},\frac{1}{3},\frac{2}{5},\frac{5}{9}\) when arranged in increasing order are :

  1. AOption (a)

    \(\frac{1}{2},\frac{1}{3},\frac{2}{5},\frac{5}{9}\)

  2. BOption (b)

    \(\frac{1}{3},\frac{1}{2},\frac{2}{5},\frac{5}{9}\)

  3. COption (c)

    \(\frac{1}{3},\frac{1}{2},\frac{5}{9},\frac{2}{5}\)

  4. DOption (d)

    \(\frac{1}{3},\frac{2}{5},\frac{1}{2},\frac{5}{9}\)

View answer & explanation

Correct answer

Answer: (d)D.

\(\frac{1}{3},\frac{2}{5},\frac{1}{2},\frac{5}{9}\)

Why this answer is correct

Calculation:

Let's find the decimal equivalents of each fraction:

1/2 = 0.5

1/3 = 0.333

2/5 = 0.4

5/9 = 0.556

Now, arranging them in increasing order:

  1. 1/3 = 0.333

  2. 2/5 = 0.4

  3. 1/2 = 0.5

  4. 5/9 = 0.556

So, the correct order is: ​1/3, 2/5, 1/2, 5/9​​ ​​

Question 157

KVS PRT 2023• Subject Specific

The product of two numbers is 1728. If their HCF is 12, Their LCM is :

  1. AOption (a)

    160

  2. BOption (b)

    144

  3. COption (c)

    136

  4. DOption (d)

    120

View answer & explanation

Correct answer

Answer: (b)B.

144

Why this answer is correct

Concept use:

We can use the relationship between the product of two numbers (N1 × N2), their highest common factor (HCF), and their least common multiple (LCM):

N1 ×  N2 = HCF ×  LCM

Calculation

Given that the product of the two numbers is 1728 and their HCF is 12, we can rearrange the equation to find the LCM:

LCM = (N1 ×  N2) / HCF

Substituting the given values:

LCM = 1728 / 12

LCM = 144

So, the LCM of the two numbers is 144.

Question 158

KVS PRT 2023• Subject Specific

1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + _ _ _ _ _ + 151 on simplification, gives :

  1. AOption (a)

    60

  2. BOption (b)

    70

  3. COption (c)

    75

  4. DOption (d)

    76

View answer & explanation

Correct answer

Answer: (d)D.

76

Why this answer is correct

Explanation:

This expression is an alternating series where each term alternates between addition and subtraction.

The pattern follows:

1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + ... + 151

To simplify this expression, we group the terms:

(1 - 2) + (3 - 4) + (5 - 6) + ... + (147 - 148) + (149 - 150) + 151

Each pair within the parentheses equals -1, and there are 75 such pairs (since 150/2 = 75). So the sum of the paired terms is -75, and we still have the final +151.

∴ ⇒ (1 - 2) + (3 - 4) + ... + (149 - 150) + 151

   ⇒ -75 + 151

   ⇒ 76

The correct answer is option 4) 76.

Question 159

KVS PRT 2023• Subject Specific

There are three ropes of length 240 cm, 320 cm and 460 cm which are to be cut into equal lengths. The greatest possible length of each piece is :

  1. AOption (a)

    80 cm

  2. BOption (b)

    60 cm

  3. COption (c)

    20 cm

  4. DOption (d)

    10 cm

View answer & explanation

Correct answer

Answer: (c)C.

20 cm

Why this answer is correct

Concept use:

The Greatest Common Divisor (GCD), also known as the Greatest Common Factor (GCF) or Highest Common Factor (HCF), is the largest positive integer that divides each of the given numbers without leaving a remainder.

Explanation:

To find the greatest possible length of each piece into which the ropes can be cut equally,

we need to find the greatest common divisor (GCD) of the lengths of the ropes.

The lengths of the ropes are 240 cm, 320 cm, and 460 cm.

Now, let's find the GCD of these lengths:

GCD(240, 320, 460) = 20

Therefore, the greatest possible length of each piece is 20 cm.

So, the correct answer is 3) 20 cm

Question 160

KVS PRT 2023• Subject Specific

The number of 2-digit prime numbers in which digits are also prime is :

  1. AOption (a)

    4

  2. BOption (b)

    6

  3. COption (c)

    8

  4. DOption (d)

    21

View answer & explanation

Correct answer

Answer: (a)A.

4

Why this answer is correct

Explanation:

To determine the number of 2-digit prime numbers in which both digits are also prime, we first list the prime numbers that have single digits. These single-digit prime numbers are: 2, 3, 5, and 7.

Now we need to form 2-digit numbers where both digits are prime and then check which of these numbers are prime.

The possible combinations are:

22, 23, 25, 27

32, 33, 35, 37

52, 53, 55, 57

72, 73, 75, 77

Next, we need to identify which of these combinations are prime numbers:

23, 37, 53 and  73  is prime

Thus, the number of 2-digit prime numbers in which both digits are also prime is: 4