CAT 2010 QA Slot 2 Question Paper(Available) :Download Solutions with Answer Key

Step 1: Understanding the problem

We are given two linear equations in two variables:

(1) \(x + 2y = 10\)

(2) \(3x - y = 5\)

We are asked to find \(x + y\). The goal is to solve the system and then sum the variables.


Step 2: Choosing a method

We can solve this system using the elimination method, which allows us to remove one variable and find the other.


Step 3: Elimination of \(y\)

From equation (2), multiply the whole equation by 2 to match the coefficient of \(y\) in equation (1):
\(2(3x - y) = 2 \cdot 5\)
\(\Rightarrow 6x - 2y = 10\) \quad (3)


Step 4: Adding equations to eliminate \(y\)

Add equation (1) and equation (3):
\((x + 2y) + (6x - 2y) = 10 + 10\)
\(7x = 20\)
\(\Rightarrow x = \frac{20}{7}\).


Step 5: Substituting back to find \(y\)

Substitute \(x = \frac{20}{7}\) into equation (1):
\(\frac{20}{7} + 2y = 10\)
\(2y = 10 - \frac{20}{7}\)
\(2y = \frac{70}{7} - \frac{20}{7} = \frac{50}{7}\)
\(\Rightarrow y = \frac{25}{7}\).


Step 6: Finding \(x + y\)
\(x + y = \frac{20}{7} + \frac{25}{7} = \frac{45}{7}\).


Step 7: Final check

Check in equation (2): \(3x - y = 3\left(\frac{20}{7}\right) - \frac{25}{7} = \frac{60}{7} - \frac{25}{7} = \frac{35}{7} = 5\), correct.


Final Answer: \(\boxed{\frac{45}{7}}\)
Quick Tip: In elimination, align coefficients to remove one variable. Always verify your answer by substituting back into both original equations.

Step 1: Define variables and known values

Let the marked price be \(M\).
Cost price (CP) = Rs. 400.
Profit percentage = 25% of CP.


Step 2: Calculate selling price (SP)

Profit = \(0.25 \times 400 = Rs. 100\).

Selling price = CP + Profit = \(400 + 100 = Rs. 500\).


Step 3: Relating SP and MP

The shopkeeper gives a discount of 20% on MP:

Discounted price = MP - (20% of MP) = \(M - 0.20M = 0.80M\).

We know this discounted price equals the SP: \(0.80M = 500\).


Step 4: Solving for M
\(M = \frac{500}{0.80} = \frac{500 \times 100}{80} = 625\).


Step 5: Verification

If MP = Rs. 625, discount = \(0.20 \times 625 = Rs. 125\), so SP = \(625 - 125 = Rs. 500\).
Profit = \(500 - 400 = Rs. 100\), which is exactly 25% of Rs. 400. Verified.


Final Answer: \(\boxed{Rs. 625}\)
Quick Tip: Link MP, SP, and CP carefully in profit-discount problems. Remember: SP = MP × (1 - discount rate).

Step 1: Understanding the problem

We are given the total distance covered by the train (240 km) and the total time taken (4 hours). We need to find the average speed in km/h.


Step 2: Formula for speed

The formula for average speed is:
\[ Speed = \frac{Distance}{Time} \]


Step 3: Substitution of values

Distance = 240 km, Time = 4 hours:
\[ Speed = \frac{240}{4} = 60 \ km/h \]


Step 4: Verification

If the speed is 60 km/h and the time is 4 hours, then distance covered = \(60 \times 4 = 240\) km, which matches the given data.


Final Answer: \(\boxed{60 \ km/h}\)
Quick Tip: Always keep distance and time in the same units when calculating speed. If the time is in minutes, convert to hours for km/h calculations.

Step 1: Represent the present ages

Let A's present age = \(3x\) years, B's present age = \(4x\) years (as per ratio 3:4).


Step 2: Write the condition for future ages

In 5 years:
A’s age = \(3x + 5\) years,
B’s age = \(4x + 5\) years.
We are told their ages will be in the ratio 4:5: \[ \frac{3x + 5}{4x + 5} = \frac{4}{5} \]


Step 3: Solve for \(x\)

Cross-multiply: \(5(3x + 5) = 4(4x + 5)\)
\(15x + 25 = 16x + 20\)
\(25 - 20 = 16x - 15x\)
\(5 = x\)


Step 4: Find A's present age

A's age = \(3x = 3 \times 5 = 15\) years.


Step 5: Verification

B's age = \(4 \times 5 = 20\) years.
In 5 years: A = 20 years, B = 25 years.
Ratio = \(20 : 25 = 4 : 5\), which matches perfectly.


Final Answer: \(\boxed{15 \ years}\)
Quick Tip: When solving ratio-age problems, always define variables proportionally to the ratio, then apply the future/past condition.

Step 1: Simplify each logarithm
\(32 = 2^5 \implies \log_{2}(32) = \log_{2}(2^5) = 5\).
\(9 = 3^2 \implies \log_{3}(9) = \log_{3}(3^2) = 2\).


Step 2: Add the results
\[ \log_{2}(32) + \log_{3}(9) = 5 + 2 = 7 \]


Step 3: Verification

Both logs are exact integers because the numbers are perfect powers of their bases.


Final Answer: \(\boxed{7}\)
Quick Tip: If the number is a perfect power of the log base, the logarithm is just the exponent.

Step 1: Understanding the problem

We are told that the money doubles in 4 years under compound interest.
We need to find the time to make it 8 times the principal.


Step 2: Finding the growth factor per 4 years

If in 4 years the amount becomes \(2P\), then: \((1+R)^4 = 2\).
Taking the fourth root: \(1+R = 2^{1/4}\).


Step 3: For 8 times the principal

We want \(P\) to become \(8P\): \((1+R)^T = 8\).
Substitute \(1+R = 2^{1/4}\): \((2^{1/4})^T = 8\).


Step 4: Express in powers of 2
\(8 = 2^3\).
So: \(2^{T/4} = 2^3\).
Equating exponents: \(\frac{T}{4} = 3 \implies T = 12\).


Final Answer: \(\boxed{12 \ years}\)
Quick Tip: In compound interest, use the doubling period to find the rate, then apply exponent rules for any growth multiple.

Step 1: Recall the area formula

Area of a circle = \(\pi r^2\).


Step 2: Substitution
\(\frac{22}{7} r^2 = 154\).


Step 3: Solve for \(r^2\)

Multiply both sides by \(\frac{7}{22}\): \(r^2 = 154 \times \frac{7}{22} = 49\).


Step 4: Find \(r\)
\(r = \sqrt{49} = 7\) cm.


Final Answer: \(\boxed{7 \ cm}\)
Quick Tip: Always keep \(\pi\) as given in the question—do not replace with 3.14 unless told.

Step 1: Apply the square identity
\((a+b)^2 = a^2 + b^2 + 2ab\).


Step 2: Substitution
\(a^2 + b^2 = 25\), \(ab = 12\): \((a+b)^2 = 25 + 2(12) = 25 + 24 = 49\).


Step 3: Solve for \(a+b\)
\(a+b = \sqrt{49} = 7\).
Since we generally consider positive sums in such contexts, we take \(+7\).


Final Answer: \(\boxed{7}\)
Quick Tip: When given \(a^2+b^2\) and \(ab\), use \((a+b)^2\) identity to avoid solving individual variables.

Step 1: Represent the speeds

Let boat speed in still water = \(B\) km/h, stream speed = \(S\) km/h.


Step 2: Downstream speed
\(B + S = \frac{60}{5} = 12\) km/h.


Step 3: Upstream speed
\(B - S = \frac{60}{6} = 10\) km/h.


Step 4: Solve for \(B\)

Add both: \((B+S) + (B-S) = 12 + 10 \implies 2B = 22 \implies B = 11\) km/h.


Final Answer: \(\boxed{11 \ km/h}\)
Quick Tip: Boat in still water = average of downstream and upstream speeds.

Step 1: Recall the AP sum formula
\(S_n = \frac{n}{2}[2a + (n-1)d]\).


Step 2: Substitution
\(670 = \frac{20}{2}[2(5) + 19d]\).
\(670 = 10[10 + 19d]\).


Step 3: Solve for \(d\)
\(67 = 10 + 19d\)
\(19d = 57\)
\(d = 3\).


Final Answer: \(\boxed{3}\)
Quick Tip: For AP, always check if given data matches an integer common difference; if not, the problem may require fractional \(d\).

Step 1: Recall the compound interest formula
\[ A = P\left(1 + \frac{R}{100}\right)^n \]
Where: \(P\) = Principal amount, \(R\) = Annual rate of interest, \(n\) = Number of years, \(A\) = Final amount.


Step 2: Substitution of values

Here: \(P = 5000\), \(R = 10%\), \(n = 2\): \[ A = 5000\left(1 + \frac{10}{100}\right)^2 = 5000(1.1)^2 \]


Step 3: Square the factor
\((1.1)^2 = 1.21\)


Step 4: Multiply with principal
\(A = 5000 \times 1.21 = 6050\)


Step 5: Verification

Year 1: \(5000 \times 1.1 = 5500\)

Year 2: \(5500 \times 1.1 = 6050\) ✔


Final Answer: \(\boxed{Rs. 6050}\)
Quick Tip: In compound interest, the interest of each year is calculated on the amount from the previous year, not just the original principal.

Step 1: From the second equation, express \(y\) in terms of \(x\)
\(x - y = 1 \implies y = x - 1\)


Step 2: Substitute \(y = x - 1\) into the first equation
\(2x + 3(x - 1) = 12\)


Step 3: Expand and simplify
\(2x + 3x - 3 = 12\)
\(5x - 3 = 12\)


Step 4: Solve for \(x\)
\(5x = 15 \implies x = 3\)


Step 5: Verification

If \(x = 3\), then \(y = 3 - 1 = 2\).
Check equation 1: \(2(3) + 3(2) = 6 + 6 = 12\) ✔

Check equation 2: \(3 - 2 = 1\) ✔


Final Answer: \(\boxed{3}\)
Quick Tip: When one equation is easily rearranged, substitution is often the quickest method for solving simultaneous equations.

Step 1: Represent dimensions with variables

Let breadth = \(B\) cm.
Length = \(B + 5\) cm (since it is 5 cm more than breadth).


Step 2: Use the perimeter formula

Perimeter \(P = 2(L + B) = 50\): \(2((B + 5) + B) = 50\)
\(2(2B + 5) = 50\)


Step 3: Solve for \(B\)
\(2B + 5 = 25\)
\(2B = 20 \implies B = 10\) cm


Step 4: Find length
\(L = B + 5 = 10 + 5 = 15\) cm


Step 5: Calculate area

Area = \(L \times B = 15 \times 10 = 150\) cm\(^2\)


Step 6: Verification

Perimeter = \(2(15 + 10) = 50\) ✔


Final Answer: \(\boxed{150 \ cm^2}\)
Quick Tip: In rectangle problems, first solve for one dimension using the perimeter, then use the area formula.

Step 1: Find each square root
\(\sqrt{169} = 13\) (since \(13^2 = 169\))
\(\sqrt{64} = 8\) (since \(8^2 = 64\))


Step 2: Add the values
\(13 + 8 = 21\)


Final Answer: \(\boxed{21}\)
Quick Tip: Memorizing perfect squares up to 20 helps to solve such problems instantly.

Step 1: Find individual work rates

Man's rate = \(\frac{1}{12}\) work/day.
Woman's rate = \(\frac{1}{18}\) work/day.


Step 2: Find combined work rate
\(\frac{1}{12} + \frac{1}{18} = \frac{3}{36} + \frac{2}{36} = \frac{5}{36}\) work/day.


Step 3: Time taken together

Time = \(\frac{1}{\frac{5}{36}} = \frac{36}{5} = 7.2\) days.


Final Answer: \(\boxed{7.2 \ days}\)
Quick Tip: When combining work rates, always add them to find the total work per day, then take the reciprocal for total time.

Step 1: Calculate profit

Profit = 25% of 200 = \(0.25 \times 200 = 50\)


Step 2: Find selling price

SP = CP + Profit = \(200 + 50 = 250\)


Final Answer: \(\boxed{Rs. 250}\)
Quick Tip: SP = CP × (1 + Profit%) is the fastest method for selling price in profit cases.

Step 1: Apply the laws of exponents
\(\frac{a^m}{a^n} = a^{m-n}\)


Step 2: Substitution
\(3^4 \div 3^2 = 3^{4-2} = 3^2\)


Step 3: Evaluate
\(3^2 = 9\)


Final Answer: \(\boxed{9}\)
Quick Tip: For same base division, subtract exponents from numerator and denominator.

Step 1: Find distances

First part: \(60 \times 3 = 180\) km.
Second part: \(80 \times 2 = 160\) km.
Total = \(180 + 160 = 340\) km.


Step 2: Find total time
\(3 + 2 = 5\) hours.


Step 3: Average speed
\(\frac{340}{5} = 68\) km/h.


Final Answer: \(\boxed{68 \ km/h}\)
Quick Tip: Average speed is total distance divided by total time, not the average of the speeds.

Step 1: Use formula
\(SP = CP \times (1 + \frac{Profit%}{100})\)


Step 2: Substitution
\(240 = CP \times 1.2\)


Step 3: Solve
\(CP = \frac{240}{1.2} = 200\)


Final Answer: \(\boxed{Rs. 200}\)
Quick Tip: To reverse profit, divide selling price by \((1 + Profit%)\).

Method 1: Prime factorization
\(48 = 2^4 \times 3\)
\(72 = 2^3 \times 3^2\)

Common factors: \(2^3\) and \(3\).
HCF = \(2^3 \times 3 = 8 \times 3 = 24\).


Method 2: Euclidean algorithm
\(72 \div 48 = 1\) remainder 24
\(48 \div 24 = 2\) remainder 0 ⇒ HCF = 24.


Final Answer: \(\boxed{24}\)
Quick Tip: Use Euclidean algorithm for faster HCF, especially with large numbers.

Step 1: Recall the definition of factorial

The factorial of a positive integer \(n\), denoted \(n!\), is the product of all positive integers from \(n\) down to \(1\). For example: \(5! = 5 \times 4 \times 3 \times 2 \times 1\)
\(4! = 4 \times 3 \times 2 \times 1\)


Step 2: Calculate \(5!\)
\(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)


Step 3: Calculate \(4!\)
\(4! = 4 \times 3 \times 2 \times 1 = 24\)


Step 4: Add the results
\(5! + 4! = 120 + 24 = 144\)


Step 5: Verification

We computed each factorial separately and summed them correctly. No simplification possible.


Final Answer: \(\boxed{144}\)
Quick Tip: When dealing with factorial expressions, always compute each factorial term separately before performing addition or subtraction.

Step 1: Understand work rate concept

Filling rate of pipe A = fraction of tank filled per hour = \(\frac{1}{6}\).
Emptying rate of pipe B = fraction of tank emptied per hour = \(\frac{1}{8}\).


Step 2: Determine net rate

Net rate = Filling rate − Emptying rate = \(\frac{1}{6} - \frac{1}{8}\).


Step 3: Subtract fractions

LCM of 6 and 8 = 24. \(\frac{1}{6} = \frac{4}{24}\), \(\frac{1}{8} = \frac{3}{24}\).
Net rate = \(\frac{4}{24} - \frac{3}{24} = \frac{1}{24}\) tank/hour.


Step 4: Find total time

If \(\frac{1}{24}\) of the tank is filled per hour, then the whole tank takes \(24\) hours.


Step 5: Verification

In 24 hours, filling pipe puts in \(\frac{24}{6} = 4\) tanks worth, emptying pipe removes \(\frac{24}{8} = 3\) tanks worth. Net = 1 tank filled. ✔


Final Answer: \(\boxed{24 \ hours}\)
Quick Tip: For problems with filling and emptying rates, express each rate as work per hour and subtract appropriately for opposite actions.

Step 1: Recall cube volume formula

Volume of a cube = \(a^3\), where \(a\) = edge length.


Step 2: Set equation
\(a^3 = 343\)


Step 3: Find cube root
\(a = \sqrt[3]{343}\)


Step 4: Simplify cube root

Since \(343 = 7 \times 7 \times 7 = 7^3\), we get \(a = 7\).


Step 5: Verification
\(7^3 = 343\), matches given volume. ✔


Final Answer: \(\boxed{7 \ cm}\)
Quick Tip: To find cube roots, factorize the number and group factors in threes.

Step 1: Recognize it's a quadratic

A quadratic equation has the form \(ax^2 + bx + c = 0\). Here \(a=1, b=-5, c=6\).


Step 2: Factorize

Find two numbers whose product = \(6\) and sum = \(-5\): \(-2\) and \(-3\).


Step 3: Write factors
\((x - 2)(x - 3) = 0\)


Step 4: Solve each factor
\(x - 2 = 0 \implies x = 2\) \(x - 3 = 0 \implies x = 3\)


Step 5: Verification

Substitute \(x=2\): \(4 - 10 + 6 = 0\) ✔
Substitute \(x=3\): \(9 - 15 + 6 = 0\) ✔


Final Answer: \(\boxed{2, 3}\)
Quick Tip: Factoring is faster than the quadratic formula when integer factors exist for \(c\).

Step 1: Sum the parts in the ratio
\(2 + 3 + 5 = 10\) parts total.


Step 2: Determine value per part

Each part = \(\frac{1000}{10} = 100\)


Step 3: Find C's share

C's share = \(5\) parts = \(5 \times 100 = 500\) Rs.


Step 4: Verification

A = \(200\), B = \(300\), C = \(500\). Sum = \(1000\). ✔


Final Answer: \(\boxed{Rs. 500}\)
Quick Tip: Always divide total sum by total parts to find per-part value in ratio problems.

Step 1: Recall trigonometric values
\(\sin 30^\circ = \frac{1}{2}\), so \(\sin^2 30^\circ = \frac{1}{4}\). \(\cos 60^\circ = \frac{1}{2}\), so \(\cos^2 60^\circ = \frac{1}{4}\).


Step 2: Add the squares
\(\frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}\).


Final Answer: \(\boxed{\frac{1}{2}}\)
Quick Tip: Memorize standard angles' sine and cosine values for quick trigonometric calculations.

Step 1: Express the number in modular form

If a number leaves remainder \(4\) when divided by \(7\), it can be written as: \[ N = 7k + 4 \]
where \(k\) is any integer.

Step 2: Square the number
\[ N^2 = (7k + 4)^2 = 49k^2 + 56k + 16 \]

Step 3: Divide each term by 7 and check remainders

- \(49k^2\) is divisible by \(7\) (remainder \(0\)).
- \(56k\) is divisible by \(7\) (remainder \(0\)).
- \(16 \div 7 = 2\) remainder \(2\).

Thus, the remainder when \(N^2\) is divided by 7 is \(2\).

Step 4: Shorter modular arithmetic method

We know: \[ N \equiv 4 \ (mod 7) \]
Squaring both sides: \[ N^2 \equiv 4^2 \ (mod 7) \implies N^2 \equiv 16 \ (mod 7) \]
Since \(16 \div 7 = 2\) remainder \(2\), \[ N^2 \equiv 2 \ (mod 7) \]

Step 5: Verification example

Let \(N = 11\) (since \(11 \div 7\) leaves remainder \(4\)). \(N^2 = 121\). \(121 \div 7 = 17\) remainder \(2\), confirming our result. ✔

Final Answer: \(\boxed{2}\)
Quick Tip: Use modular arithmetic for remainder problems — it simplifies calculations without expanding large numbers.

Step 1: Define the two numbers

Let the numbers be \(x\) and \(y\).
We are told: \[ x + y = 15 \quad (Sum) \] \[ xy = 56 \quad (Product) \]

Step 2: Form a quadratic equation

If \(x\) and \(y\) are roots of a quadratic equation, it can be written as: \[ t^2 - (x+y)t + xy = 0 \]
Substitute the given values: \[ t^2 - 15t + 56 = 0 \]

Step 3: Solve the quadratic by factorization

Find two numbers whose sum is \(-15\) (coefficient of \(t\)) and whose product is \(56\):
These are \(-7\) and \(-8\).

Factorize: \[ t^2 - 15t + 56 = (t - 7)(t - 8) = 0 \]

Step 4: Find the two numbers
\(t = 7\) or \(t = 8\).
Thus, the two numbers are \(7\) and \(8\).

Step 5: Identify the larger number

Larger = \(8\).

Step 6: Verification

Sum: \(7 + 8 = 15\) ✔
Product: \(7 \times 8 = 56\) ✔

Final Answer: \(\boxed{8}\)
Quick Tip: For sum and product problems, convert them into a quadratic equation and solve to get both numbers easily.

Step 1: Calculate the speed of the train

Speed = \(\frac{Distance}{Time} = \frac{300}{5} = 60\) km/h.

Step 2: Use the speed to find the required time

Time = \(\frac{Distance}{Speed}\) \[ Time = \frac{180}{60} = 3 \ hours \]

Step 3: Verification

If the speed is 60 km/h, then in 3 hours:
Distance = \(60 \times 3 = 180\) km, matches the given distance. ✔

Step 4: Relation between distance and time at constant speed

Since speed is constant, time is directly proportional to distance.
Here: \(\frac{180}{300} = 0.6\), and \(0.6 \times 5\) hours = 3 hours.

Final Answer: \(\boxed{3 \ hours}\)
Quick Tip: At constant speed, \(\frac{Distance_1}{Time_1} = \frac{Distance_2}{Time_2}\). Use this proportion to solve quickly.

Step 1: Multiply the numerators together

Numerator = \(2 \times 3 \times 4 = 24\).

Step 2: Multiply the denominators together

Denominator = \(3 \times 4 \times 5 = 60\).

Step 3: Form the fraction
\[ \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{24}{60} \]

Step 4: Simplify the fraction

Divide numerator and denominator by their GCD (12): \[ \frac{24}{60} = \frac{2}{5} \]

Step 5: Verification by cancellation before multiplication

Cancel common terms early: \(\frac{2}{\cancel{3}} \times \frac{\cancel{3}}{4} \times \frac{4}{5}\)
Cancel 4 with denominator 4: \(= \frac{2}{1} \times \frac{1}{1} \times \frac{1}{5} = \frac{2}{5}\) ✔

Final Answer: \(\boxed{\frac{2}{5}}\)
Quick Tip: In fraction multiplication, always simplify by cancelling common factors before multiplying to reduce calculations.

Step 1: Use the formula for sum from average

We know: \[ Sum of terms = Average \times Number of terms \]
For 5 numbers: \[ Original sum = 20 \times 5 = 100 \]

Step 2: Find the new sum after removal

After removing one number, there are 4 numbers with an average of 18: \[ New sum = 18 \times 4 = 72 \]

Step 3: Find the removed number

Removed number = Original sum − New sum: \[ Removed number = 100 - 72 = 28 \]

Step 4: Verification

If removed number is 28, then remaining sum = \(100 - 28 = 72\)
Average = \(72 \div 4 = 18\), matches perfectly. ✔

Final Answer: \(\boxed{28}\)
Quick Tip: When an element is removed from an average problem, compute the total sums before and after removal to find the missing number.

Step 1: Calculate the discount amount
\[ Discount = \frac{Discount %}{100} \times Marked Price \] \[ Discount = \frac{10}{100} \times 500 = 50 \]

Step 2: Find the selling price

Selling Price (SP) = Marked Price − Discount: \[ SP = 500 - 50 = 450 \]

Step 3: Alternative percentage method
\[ SP = MP \times \left(1 - \frac{Discount %}{100}\right) \] \[ SP = 500 \times 0.9 = 450 \]

Step 4: Verification

Discount given = Rs. 50, selling price = Rs. 450, matches both methods. ✔

Final Answer: \(\boxed{Rs. 450}\)
Quick Tip: For discount problems, remember \(SP = MP \times (1 - \frac{Discount%}{100})\) for quick calculations.

Step 1: Express \(x\) in terms of \(y\)

From \(3x = 4y\): \[ x = \frac{4y}{3} \]

Step 2: Express \(y\) in terms of \(z\)

From \(y = 2z\), substitute into \(x\): \[ x = \frac{4(2z)}{3} = \frac{8z}{3} \]
Thus: \[ x : y : z = \frac{8z}{3} : 2z : z \]

Step 3: Remove the fraction

Multiply each term by 3: \[ 8z : 6z : 3z \]

Step 4: Simplify the ratio

Cancel \(z\) from each term: \[ 8 : 6 : 3 \]

Step 5: Verification

Check with actual values: Let \(z=3\), then \(y=6\), \(x=\frac{8(3)}{3}=8\). Ratio = \(8:6:3\), correct. ✔

Final Answer: \(\boxed{8:6:3}\)
Quick Tip: When given equations relating variables, express all in terms of one variable to easily find ratios.

Step 1: Recall the meaning of exponent
\(2^{10}\) means multiplying \(2\) by itself \(10\) times: \[ 2^{10} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \]

Step 2: Break down into smaller powers

We know: \(2^5 = 32\).
Thus: \[ 2^{10} = (2^5)^2 = 32^2 \]

Step 3: Square 32
\[ 32^2 = 1024 \]

Step 4: Verification

Multiply step-by-step: \(2^2 = 4\), \(2^3 = 8\), \(2^4 = 16\), \(2^5 = 32\), \(2^{10} = 32 \times 32 = 1024\). ✔

Final Answer: \(\boxed{1024}\)
Quick Tip: When calculating high powers, break them into known powers for easier multiplication.