Explanation : 

Number of zeroes at the end of product

1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60

= 1 × 5 × (5 × 2) × (5 × 3) × (5 × 2^2) × (5 × 5) × (5 × 3 × 2) × (5 × 7) × (5 × 2^3) × (5 × 9) × (5 × 5 × 2) × (5 × 11) × (5 × 2^2 × 3)

Here, number of 2s = 10
And number of 5s = 14
The lesser of the two will determine the number of zeros.
Hence, there will be 10 zeros at the end in the given expression.

Model Answer:

If you have two straight sticks of length 7.5 feet and 3.25 feet

To find the minimum length that can be measured using two straight sticks of length 7.5 feet and 3.25 feet using the Highest Common Factor (HCF) method, we can first convert the lengths to inches:

7.5 feet = 90 inches
3.25 feet = 39 inches

Then, we can find the HCF of 90 and 39:

90 = 2 * 3 * 3 * 5
39 = 3 * 13

The common factor is 3, so the HCF is 3 inches.

Therefore, the minimum length that can be measured with these two sticks is 3 inches, which is equivalent to 0.25 feet.

Hence, the correct answer is (b) 0.25 foot.

Explanation : 

As a result of 25 hike -> Finding original price of rice 

Given Information:
– Original price of rice = x rupees per kg
– Price increased by 25%
– New price = 1.25x rupees per kg
– Available money = Rs. 1,200
– Difference in quantity = 6 kg

Step 1: Calculate quantity before price hike
Original quantity = 1200/x kg

Step 2: Calculate quantity after price hike
New quantity = 1200/(1.25x) kg

Step 3: Set up equation using difference
1200/x – 1200/(1.25x) = 6

Step 4: Solve for x
1200/x – 960/x = 6
(1200 – 960)/x = 6
240/x = 6
x = 40

Therefore:
– Original price of rice = Rs. 40 per kg
– After 25% increase, new price = Rs. 50 per kg
– Original quantity = 1200/40 = 30 kg
– New quantity = 1200/50 = 24 kg
– Difference = 30 – 24 = 6 kg

Explanation : 

All cats are dogs. All cats are black.

1. For Conclusion I – “All dogs are black”:
– While we know all cats are black and all cats are dogs
– We cannot conclude all dogs are black
– Because not all dogs may be cats
– Example: There could be dogs that are not cats, and these dogs could be any color

2. For Conclusion II – “Some dogs are not black”:
– We cannot conclude this either
– The statements give no information about dogs that are not cats
– Making any assumption about non-cat dogs would be illogical

Therefore:
Neither conclusion can be logically derived from the given statements. The statements only tell us about cats (that they are dogs and they are black) but give no complete information about dogs.

Correct Answer: Neither Conclusion-I nor Conclusion-II follows from the given statements.

Explanation : 

A is 16th from the left end in a row

Problem: Finding total number of boys in a row

Given:
– A is 16th from left end
– V is 18th from right end
– Q is 11th from A towards right
– Q is 3rd from V towards right

Solution:
1. Find Q’s position from left:
– A is 16th from left
– Q is 11th right of A
– So Q is 16 + 11 = 27th from left

2. Find Q’s position from right:
– V is 18th from right
– Q is 3rd right of V
– So Q is 18 – 3 = 15th from right

3. Calculate total boys:
– Q’s position from left = 27
– Q’s position from right = 15
– Total boys = Left position + Right position – 1
– Total = 27 + 15 – 1 = 41 (Subtract 1 because Q is counted twice)

Therefore, there are 41 boys in the row.

Explanation :

Problem: Find pairs of numbers whose difference of squares is 63

Given:
– Two natural numbers x and y where x > y
– x² – y² = 63

Solution:
1. Factor the equation:
x² – y² = (x+y)(x-y) = 63

2. Find factors of 63:
63 = 1 × 63
63 = 3 × 21
63 = 7 × 9

3. For each factor pair (a,b):
x + y = larger number
x – y = smaller number

Solving:
x = (larger + smaller)/2
y = (larger – smaller)/2

4. Calculate pairs:
For 63,1:
x = (63 + 1)/2 = 32
y = (63 – 1)/2 = 31

For 21,3:
x = (21 + 3)/2 = 12
y = (21 – 3)/2 = 9

For 9,7:
x = (9 + 7)/2 = 8
y = (9 – 7)/2 = 1

Therefore, three pairs of numbers exist: (32,31), (12,9), and (8,1)

Explanation : 

Problem: Finding the day of week in July if 12th January is a Sunday

Given:
– 12th January is a Sunday
– Need to determine if 12th July is a Sunday in a leap year

Solution:
1. Count days from 12th January to 12th July:
– Remaining days in January = 31 – 12 = 19 days
– February (leap year) = 29 days
– March = 31 days
– April = 30 days
– May = 31 days
– June = 30 days
– July (till 12th) = 12 days

2. Total days = 19 + 29 + 31 + 30 + 31 + 30 + 12 = 182 days

3. Calculate odd days:
– 182 ÷ 7 = 26 weeks + 0 days
– 0 odd days means same day as start

4. Since 12th January was Sunday:
– 0 odd days means 12th July will also be Sunday

Therefore, in a leap year, 12th July will be a Sunday.

Explanation :

Frog tries to come out of well

Given:
– Well depth = 4.5 meters = 450 cm
– Each jump up = 30 cm
– Each slide down = 15 cm
– Net progress per jump = 30 cm – 15 cm = 15 cm

Solution:
1. Initial calculation:
– Distance to cover = 450 cm
– Net progress per jump = 15 cm
– Number of jumps = 450 ÷ 15 = 30 jumps

2. Important consideration:
– When frog reaches the top, it won’t slide down
– Last jump will give full 30 cm progress, not 15 cm
– This saves one jump

3. Final calculation:
– First 28 jumps: 28 × 15 cm = 420 cm
– Last jump: 30 cm
– Total distance = 420 + 30 = 450 cm
– Total jumps needed = 29

Therefore, the frog needs 29 jumps to come out of the well.

Explanation :

Let’s solve how long it takes two people to complete 40% of work together.

Given:
Person 1:
– Can do 20% (W/5) of work in 8 days
– Rate = (W/5)/8 = W/40 per day

Person 2:
– Can do 25% (W/4) of work in 6 days
– Rate = (W/4)/6 = W/24 per day

Step by step calculations:
1. Combined rate of work = Rate of Person 1 + Rate of Person 2
= W/40 + W/24
= (24W + 40W)/(40 × 24)
= 64W/960
= W/15 per day

2. Time to complete 40% work:
Work = Rate × Time
0.4W = (W/15) × Time
Time = (0.4W) ÷ (W/15)
= 0.4W × (15/W)
= 0.4 × 15
= 6 days

Therefore, working together, they will complete 40% of the work in 6 days.

Explanation :

In a class there are three groups a b and c ….

1. Average weight = Total weight ÷ Number of students

2. When students are shifted between groups:
– Total weight of all students remains the same
– Total number of students remains the same
– Students are just reorganized within the same class

Example:
Let’s say in a class of 40 students:
– Total weight = 2000 kg
– Average weight = 2000 ÷ 40 = 50 kg

If students are divided into 2 groups:
– Whether 20-20 students
– Or 15-25 students
– Or any other combination
The class average will still be 50 kg because:
– Total weight is still 2000 kg
– Number of students is still 40

Therefore, shifting students between groups within the same class doesn’t affect the overall average weight.