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Basic Numeracy

295 questions

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Question 21Question →

What is the unit digit of the expression 820+7138^{20} + 7^{13}?

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Answer: 3

Answer

The unit digit of the given expression is 3.
The unit digit of 8208^{20} is 6 because 20 is a multiple of 4, matching the 4th term in the cyclicity cycle (8, 4, 2, 6). The unit digit of 7137^{13} is 7 because 13 leaves a remainder of 1 when divided by 4, matching the 1st term in the cyclicity cycle (7, 9, 3, 1). Adding these unit digits gives 6+7=136 + 7 = 13, whose unit digit is 3.

Step-by-Step Solution

1
Find the unit digit of 8208^{20}
Unit digit is 6
The cyclicity of numbers ending in 8 is 4 (pattern: 8, 4, 2, 6). Since 20 is divisible by 4 (20(mod4)=020 \pmod 4 = 0), the unit digit corresponds to the 4th power in the cycle, which is 6.
2
Find the unit digit of 7137^{13}
Unit digit is 7
The cyclicity of numbers ending in 7 is 4 (pattern: 7, 9, 3, 1). Dividing 13 by 4 gives a remainder of 1 (13(mod4)=113 \pmod 4 = 1), so the unit digit corresponds to 71=77^1 = 7.
3
Add the unit digits of both terms
3
Sum of unit digits is 6+7=136 + 7 = 13. Taking the unit digit of 13 gives 3.

Key Concept

Cyclicity of Unit Digits
Estimated Time:45s
Question 22Question →
What is the simplified value of the following mathematical expression when evaluated strictly using the BODMAS rule?
25% of 160+[45÷{15−(3.5×2+8−5‾)}]×2.525\% \text{ of } 160 + \left[ 45 \div \left\{ 15 - \left( 3.5 \times 2 + \overline{8 - 5} \right) \right\} \right] \times 2.5
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Answer: 62.5

Answer

62.5
Evaluating step-by-step according to BODMAS: first the vinculum gives 3; the round bracket gives 3.5 × 2 + 3 = 10; the curly bracket gives 15 - 10 = 5; the square bracket gives 45 ÷ 5 = 9. Next, 25% of 160 yields 40. Then multiplication gives 9 × 2.5 = 22.5. Finally, adding 40 + 22.5 gives 62.5.

Step-by-Step Solution

1
Evaluate the expression under the bar (vinculum)
\overline{8 - 5} = 3
The bar (vinculum) acts as the innermost bracket and takes highest priority.
2
Evaluate the innermost round brackets, applying multiplication before addition
(3.5 \times 2 + 3) = (7 + 3) = 10
Within round brackets, multiplication precedes addition.
3
Evaluate the curly brackets
\{15 - 10\} = 5
Perform subtraction inside the curly brackets.
4
Evaluate the square brackets
[45÷5]=9[45 \div 5] = 9
Perform division inside the square brackets.
5
Calculate the percentage term ('of' operation)
25\% \text{ of } 160 = \frac{25}{100} \times 160 = 40
'Of' represents multiplication linked to percentage.
6
Perform multiplication before addition in the outer expression
9×2.5=22.59 \times 2.5 = 22.5
Multiplication takes precedence over addition.
7
Perform final addition
40 + 22.5 = 62.5
Add the result of the percentage term to the bracket evaluation term.

Key Concept

BODMAS Rule (Brackets, Orders, Division/Multiplication, Addition/Subtraction) with Vinculum and Percentages
Question 23Question →
What is the simplified value of the following mathematical expression when evaluated using the standard BODMAS rule?
25% of 180−[4.8+{12÷(313×0.8+0.4‾)}]25\% \text{ of } 180 - \left[ 4.8 + \left\{ 12 \div \left( 3\frac{1}{3} \times \overline{0.8 + 0.4} \right) \right\} \right]
Show answer & explanation

Answer: 37.237.2

Answer

The simplified value of the expression is 37.237.2.
Evaluating strictly according to the BODMAS hierarchical priority: first resolve the vinculum 0.8+0.4‾=1.2\overline{0.8 + 0.4} = 1.2, then the round brackets 103×1.2=4\frac{10}{3} \times 1.2 = 4, then the division in curly brackets 12÷4=312 \div 4 = 3, followed by square bracket addition 4.8+3=7.84.8 + 3 = 7.8, and finally subtracting this from 25% of 180=4525\% \text{ of } 180 = 45 gives 45−7.8=37.245 - 7.8 = 37.2.

Step-by-Step Solution

1
Evaluate the expression under the vinculum bar first
0.8+0.4‾=1.2\overline{0.8 + 0.4} = 1.2
According to the BODMAS rule, operations under a vinculum bar have the highest priority over all other brackets.
2
Convert the mixed fraction and evaluate the round brackets
313×1.2=103×1210=43\frac{1}{3} \times 1.2 = \frac{10}{3} \times \frac{12}{10} = 4
Convert 3133\frac{1}{3} to the improper fraction 103\frac{10}{3} and multiply by 1.21.2 (or 65\frac{6}{5}).
3
Evaluate the expression inside the curly brackets
12÷4=312 \div 4 = 3
Perform division of 1212 by the result of the round brackets (44).
4
Evaluate the expression inside the square brackets
4.8+3=7.84.8 + 3 = 7.8
Add 4.84.8 to the result obtained from the curly brackets (33).
5
Calculate the percentage term and perform the final subtraction
25% of 180=25100×180=4525\% \text{ of } 180 = \frac{25}{100} \times 180 = 45; then 45−7.8=37.245 - 7.8 = 37.2
'Of' represents multiplication for the percentage term, which is evaluated and then reduced by the square bracket result.

Key Concept

BODMAS Rule with Vinculum and Mixed Operations
Question 24Question →

What is the unit digit of the composite exponential expression S=717100+424101−818100S = 717^{100} + 424^{101} - 818^{100}?

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Answer: 9

Answer

The unit digit of the expression is 9.
Evaluating each term using its base cyclicity gives unit digits of 1 for 717100717^{100}, 4 for 424101424^{101}, and 6 for 818100818^{100}. Combining them yields 1+4−6=−11 + 4 - 6 = -1. Adding 10 to adjust for borrowing in unit digit subtraction gives −1+10=9-1 + 10 = 9.

Step-by-Step Solution

1
Find the unit digit of 717100717^{100}
Unit digit is 1
The unit digit of the base is 7. The cyclicity of 7 is 4 (71=7,72=9,73=3,74=17^1=7, 7^2=9, 7^3=3, 7^4=1). Dividing the exponent 100 by 4 leaves remainder 0, which corresponds to the 4th power: 74≡1(mod10)7^4 \equiv 1 \pmod{10}.
2
Find the unit digit of 424101424^{101}
Unit digit is 4
The unit digit of the base is 4. The cyclicity of 4 is 2 (4odd=4,4even=64^{\text{odd}}=4, 4^{\text{even}}=6). Since the exponent 101 is odd, the unit digit is 4.
3
Find the unit digit of 818100818^{100}
Unit digit is 6
The unit digit of the base is 8. The cyclicity of 8 is 4 (81=8,82=4,83=2,84=68^1=8, 8^2=4, 8^3=2, 8^4=6). Dividing the exponent 100 by 4 leaves remainder 0, which corresponds to the 4th power: 84≡6(mod10)8^4 \equiv 6 \pmod{10}.
4
Combine the unit digits to find the final result
Unit digit is 9
Substitute the individual unit digits into the expression: 1+4−6=5−6=−11 + 4 - 6 = 5 - 6 = -1. Converting −1-1 to a positive unit digit modulo 10 gives −1+10=9-1 + 10 = 9.

Key Concept

Unit digit determination using cyclicity and modular arithmetic for composite exponential expressions.
Estimated Time:1m 15s
Question 25Question →
Evaluate the following complex mathematical expression strictly adhering to the hierarchical order of operations (BODMAS rule):
{64% of 125+[133÷{216−(1.4×2.5−35+710‾)}]}×0.375\left\{ 64\% \text{ of } 125 + \left[ \frac{13}{3} \div \left\{ 2\frac{1}{6} - \left( 1.4 \times 2.5 - \overline{\frac{3}{5} + \frac{7}{10}} \right) \right\} \right] \right\} \times 0.375
What is the exact numerical value of the final simplified expression?
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Answer: -18.75

Answer

The simplified final value of the expression is -18.75.
Following the BODMAS rule strictly: first simplify under the vinculum (3/5 + 7/10 = 1.3), then inside the round brackets (1.4 * 2.5 - 1.3 = 2.2). Next, inside the curly brackets (13/6 - 2.2 = -1/30), then the division in the square brackets ((13/3) / (-1/30) = -130). Evaluating 64% of 125 gives 80. Combining inside the main bracket gives 80 + (-130) = -50. Finally, multiplying by 0.375 yields -18.75.

Step-by-Step Solution

1
Evaluate the expression under the vinculum bar
35+710=1.3\frac{3}{5} + \frac{7}{10} = 1.3
The vinculum (bar bracket) has the highest priority under BODMAS and must be evaluated first.
2
Simplify the innermost round brackets
1.4 \times 2.5 - 1.3 = 3.5 - 1.3 = 2.2
Perform multiplication inside the round brackets before subtraction, then subtract the vinculum result.
3
Evaluate the expression inside the inner curly brackets
2\frac{1}{6} - 2.2 = \frac{13}{6} - \frac{11}{5} = -\frac{1}{30}
Convert the mixed fraction and decimal to improper fractions and subtract.
4
Evaluate the square bracket division
\frac{13}{3} \div \left( -\frac{1}{30} \right) = \frac{13}{3} \times (-30) = -130
Dividing by a fraction is equivalent to multiplying by its reciprocal.
5
Calculate the percentage term '64% of 125'
64100×125=80\frac{64}{100} \times 125 = 80
'Of' operation takes precedence over addition.
6
Combine the terms within the main outer brackets
80 + (-130) = -50
Add the result of the percentage term to the square bracket result.
7
Multiply by the outer factor 0.375
−50×0.375=−18.75-50 \times 0.375 = -18.75
Perform final multiplication to complete the simplification.

Key Concept

BODMAS order of operations with nested brackets, vinculum, fractions, decimals, and percentage operations.
Estimated Time:3m 0s
Question 26Question →

Find the unit digit of the numerical expression N=(56763×23445)−34337N = (567^{63} \times 234^{45}) - 343^{37}.

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Answer: 9

Answer

9
The unit digit of 56763567^{63} is 3 and for 23445234^{45} it is 4, making their product's unit digit 2. The unit digit of 34337343^{37} is 3. Subtracting 3 from 2 with regrouping (12−312 - 3) yields 9.

Step-by-Step Solution

1
Determine the unit digit of 56763567^{63}
3
Base unit digit is 7 with cyclicity 4. Since 63(mod4)=363 \pmod 4 = 3, 737^3 gives a unit digit of 3.
2
Determine the unit digit of 23445234^{45}
4
Base unit digit is 4 with cyclicity 2. An odd exponent yields a unit digit of 4.
3
Multiply the unit digits of the first two terms
2
The unit digit of the product is (3×4)(mod10)=2(3 \times 4) \pmod{10} = 2.
4
Determine the unit digit of 34337343^{37}
3
Base unit digit is 3 with cyclicity 4. Since 37(mod4)=137 \pmod 4 = 1, 313^1 gives a unit digit of 3.
5
Subtract the unit digit of the second part from the first part
9
Subtracting 3 from 2 requires borrowing 10 (12−3=912 - 3 = 9) to yield a valid positive unit digit.

Key Concept

Unit Digit and Cyclicity of Powers
Question 27Question →
What is the simplified value of the following mathematical expression when evaluated using the standard BODMAS rule?
60−[15+{20÷5+3×(8−4+2‾)}]60 - [15 + \{20 \div 5 + 3 \times (8 - \overline{4 + 2})\}]
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Answer: 3535

Answer

The simplified value of the expression is 3535.
Following the BODMAS rule strictly: first resolve the vinculum 4+2‾=6\overline{4 + 2} = 6, then the round brackets (8−6)=2(8 - 6) = 2. Inside the curly brackets, division and multiplication are evaluated before addition (20÷5+3×2=4+6=1020 \div 5 + 3 \times 2 = 4 + 6 = 10). Adding this to 1515 inside the square brackets gives 2525. Subtracting 2525 from 6060 gives the correct simplified result of 3535.

Step-by-Step Solution

1
Evaluate the expression under the vinculum bar.
4+2‾=6\overline{4 + 2} = 6
Operations under a vinculum (line bracket) take highest priority.
2
Simplify the innermost round brackets (8−6)(8 - 6).
8−6=28 - 6 = 2
Parentheses must be evaluated next.
3
Evaluate the terms inside the curly brackets {20÷5+3×2}\{20 \div 5 + 3 \times 2\}. Perform division and multiplication before addition.
20÷5=420 \div 5 = 4 and 3×2=63 \times 2 = 6, so 4+6=104 + 6 = 10
According to BODMAS, division and multiplication precede addition.
4
Simplify the square brackets [15+10][15 + 10].
15+10=2515 + 10 = 25
Resolve the remaining bracket structure.
5
Perform the final subtraction 60−2560 - 25.
3535
Complete the outermost operation.

Key Concept

Simplification using hierarchical order of operations (BODMAS / Vinculum rule)
Question 28Question →

Three automatic signaling beacons in a traffic control system flash at regular time intervals of 415\frac{4}{15} hours, 920\frac{9}{20} hours, and 1435\frac{14}{35} hours, respectively. If all three beacons flash simultaneously at 12:00 noon, after how many hours will they all flash together again for the first time?

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Answer: 7.2

Answer

7.2 hours
To find when events occurring at fractional time intervals coincide, compute the LCM of the fractions. Crucially, each fraction must first be simplified to its lowest terms. 1435\frac{14}{35} reduces to 25\frac{2}{5}. Taking LCM(4,9,2)HCF(15,20,5)\frac{\text{LCM}(4, 9, 2)}{\text{HCF}(15, 20, 5)} gives 365=7.2\frac{36}{5} = 7.2 hours.

Step-by-Step Solution

1
Simplify all given fractional time intervals to their lowest terms.
The reduced fractions are 415\frac{4}{15}, 920\frac{9}{20}, and 25\frac{2}{5}.
The standard LCM formula for fractions LCM of numeratorsHCF of denominators\frac{\text{LCM of numerators}}{\text{HCF of denominators}} is mathematically valid only when all fractions are reduced to co-prime numerator-denominator pairs.
2
Calculate the LCM of the numerators.
LCM(4,9,2)=36\text{LCM}(4, 9, 2) = 36.
The least common multiple of 4=224 = 2^2, 9=329 = 3^2, and 2=212 = 2^1 is 22×32=362^2 \times 3^2 = 36.
3
Calculate the HCF of the denominators.
HCF(15,20,5)=5\text{HCF}(15, 20, 5) = 5.
The highest common factor dividing 1515, 2020, and 55 is 55.
4
Divide the numerator LCM by the denominator HCF to find the simultaneous flashing interval.
365=7.2\frac{36}{5} = 7.2 hours.
The LCM of the fractional time intervals determines the minimum duration before all events synchronize.

Key Concept

LCM of Fractions with Mandatory Simplification
Estimated Time:2m 0s
Question 29Question →

What is the unit digit of the composite exponential expression E=(238440+547321)×929103E = (238^{440} + 547^{321}) \times 929^{103}?

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Answer: 7

Answer

The unit digit of the given expression is 7.
The unit digit of 238440238^{440} is 6 (since 440(mod4)=0  ⟹  84→6440 \pmod 4 = 0 \implies 8^4 \rightarrow 6). The unit digit of 547321547^{321} is 7 (since 321(mod4)=1  ⟹  71→7321 \pmod 4 = 1 \implies 7^1 \rightarrow 7). Their sum has a unit digit of (6+7)→3(6 + 7) \rightarrow 3. The unit digit of 929103929^{103} is 9 (since 103 is odd   ⟹  91→9\implies 9^1 \rightarrow 9). Multiplying the components gives 3×9=273 \times 9 = 27, yielding a final unit digit of 7.

Step-by-Step Solution

1
Find the unit digit of 238440238^{440}.
The unit digit of base 238 is 8. The cyclicity of 8 is 4 (8,4,2,68, 4, 2, 6). Divide the exponent 440 by 4: 440(mod4)=0440 \pmod 4 = 0. Since the remainder is 0, we take the 4th power in the cycle (848^4), which ends in 6.
When an exponent is completely divisible by the cycle length (remainder 0), the position corresponds to the maximum cycle power (power 4).
2
Find the unit digit of 547321547^{321}.
The unit digit of base 547 is 7. The cyclicity of 7 is 4 (7,9,3,17, 9, 3, 1). Divide exponent 321 by 4: 321(mod4)=1321 \pmod 4 = 1. The 1st power in the cycle (717^1) ends in 7.
The unit digit is determined by the remainder of the exponent modulo 4.
3
Calculate the unit digit of the expression inside the parentheses: (238440+547321)(238^{440} + 547^{321}).
Unit digit = (6+7)(mod10)=13(mod10)=3(6 + 7) \pmod{10} = 13 \pmod{10} = 3.
Addition of unit digits follows standard modular arithmetic.
4
Find the unit digit of 929103929^{103}.
The unit digit of base 929 is 9. The cyclicity of 9 is 2 (9,19, 1). Since the exponent 103 is odd, 9odd9^{\text{odd}} ends in 9.
Bases ending in 9 have an alternating cyclicity of 2.
5
Compute the final unit digit of E=(3×9)E = (3 \times 9).
Unit digit = (3×9)(mod10)=27(mod10)=7(3 \times 9) \pmod{10} = 27 \pmod{10} = 7.
Multiplying the resultant unit digits gives the final unit digit of the composite product.

Key Concept

Unit digit cyclicity rule and exponent modulo operations
Estimated Time:2m 0s
Question 30Question →

What is the simplified value of the following mathematical expression when evaluated using the standard BODMAS rule?

20+4×[36÷{12−(3+7−4‾)}]20 + 4 \times \left[ 36 \div \left\{ 12 - \left( 3 + \overline{7 - 4} \right) \right\} \right]
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Answer: 44

Answer

44
Following the BODMAS order of operations, we first evaluate the vinculum line to get 3, then solve the round bracket (3 + 3 = 6), followed by the curly bracket (12 - 6 = 6), and then the square bracket (36 ÷ 6 = 6). Finally, performing multiplication before addition gives 20 + (4 × 6) = 20 + 24 = 44.

Step-by-Step Solution

1
Simplify the expression under the vinculum (bar)
7 - 4 = 3
The vinculum has the highest priority and must be evaluated first.
2
Simplify the innermost round brackets
3 + 3 = 6
Perform addition inside the parentheses.
3
Simplify the curly brackets
12 - 6 = 6
Subtract the result of the round brackets from 12.
4
Simplify the square brackets
36 ÷ 6 = 6
Perform division inside the square brackets.
5
Apply multiplication before addition in the outer expression
4 × 6 = 24
Multiplication has higher priority than addition according to BODMAS.
6
Perform final addition
20 + 24 = 44
Add 20 to 24 to get the final simplified answer.

Key Concept

Simplification and BODMAS Rule
Question 31Question →
What is the simplified numerical value of the following mathematical expression when evaluated using the correct BODMAS order of operations?
20% of 150−[3.5+2×{7.2−(1.4+3.8−1.6‾)÷0.6}]20\% \text{ of } 150 - \left[ 3.5 + 2 \times \left\{ 7.2 - \left( 1.4 + \overline{3.8 - 1.6} \right) \div 0.6 \right\} \right]
Show answer & explanation

Answer: 24.1

Answer

24.1
Evaluating the expression by strictly following the VBODMAS hierarchy (Vinculum -> Brackets -> Of/Orders -> Division -> Multiplication -> Addition -> Subtraction) yields the step-by-step values: vinculum gives 2.2, round brackets give 3.6, division inside curly braces gives 6, curly braces give 1.2, multiplication inside square brackets gives 2.4, square brackets give 5.9, percentage gives 30, and final subtraction gives 24.1.

Step-by-Step Solution

1
Evaluate the expression under the vinculum (bar)
\overline{3.8 - 1.6} = 2.2
The vinculum has the highest priority in the hierarchy of brackets.
2
Simplify the terms within the parentheses (round brackets)
1.4 + 2.2 = 3.6
Complete the inner addition inside the round brackets.
3
Perform division inside the braces (curly brackets)
3.6÷0.6=63.6 \div 0.6 = 6
Division takes precedence over subtraction inside the curly brackets.
4
Perform subtraction inside the braces (curly brackets)
7.2 - 6 = 1.2
Complete the remaining operation inside the curly brackets.
5
Perform multiplication inside the square brackets
2×1.2=2.42 \times 1.2 = 2.4
Multiplication takes precedence over addition inside the square brackets.
6
Perform addition inside the square brackets
3.5 + 2.4 = 5.9
Complete the final operation inside the square brackets.
7
Calculate the percentage value
20\% \text{ of } 150 = \frac{20}{100} \times 150 = 30
Evaluate the 'Of' operation / percentage computation.
8
Subtract the bracket result from the percentage value
30 - 5.9 = 24.1
Execute the final subtraction outside all brackets.

Key Concept

Hierarchy of Order of Operations (BODMAS / VBODMAS)
Question 32Question →

If x=7+43x = \sqrt{7 + 4\sqrt{3}}, y=7−43y = \sqrt{7 - 4\sqrt{3}}, and z=2+53+2−53z = \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}, what is the value of the expression x3+y3x2+y2−z\frac{x^3 + y^3}{x^2 + y^2 - z}?

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Answer: 4

Answer

4
Expressing 7±437 \pm 4\sqrt{3} as perfect squares (2±3)2(2 \pm \sqrt{3})^2 simplifies xx to 2+32 + \sqrt{3} and yy to 2−32 - \sqrt{3}. This yields x+y=4x+y=4, xy=1xy=1, x2+y2=14x^2+y^2=14, and x3+y3=52x^3+y^3=52. For zz, using the cubic identity z3=a3+b3+3abzz^3 = a^3 + b^3 + 3ab z transforms the expression into z3+3z−4=0z^3 + 3z - 4 = 0, yielding the real solution z=1z = 1. Substituting these values gives 5214−1=4\frac{52}{14 - 1} = 4.

Step-by-Step Solution

1
Simplify the square root surds xx and yy
x=2+3x = 2 + \sqrt{3} and y=2−3y = 2 - \sqrt{3}
Since 7+43=4+3+2(2)(3)=(2+3)27 + 4\sqrt{3} = 4 + 3 + 2(2)(\sqrt{3}) = (2 + \sqrt{3})^2, taking the square root gives 2+32 + \sqrt{3}. Similarly, 7−43=(2−3)27 - 4\sqrt{3} = (2 - \sqrt{3})^2.
2
Calculate fundamental algebraic combinations of xx and yy
x+y=4x + y = 4, xy=1xy = 1, x2+y2=14x^2 + y^2 = 14, and x3+y3=52x^3 + y^3 = 52
x+y=(2+3)+(2−3)=4x + y = (2+\sqrt{3}) + (2-\sqrt{3}) = 4. xy=(2+3)(2−3)=4−3=1xy = (2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1. x2+y2=(x+y)2−2xy=16−2=14x^2 + y^2 = (x+y)^2 - 2xy = 16 - 2 = 14. x3+y3=(x+y)3−3xy(x+y)=64−3(1)(4)=52x^3 + y^3 = (x+y)^3 - 3xy(x+y) = 64 - 3(1)(4) = 52.
3
Evaluate the nested cube root expression for zz
z=1z = 1
Let z=a+bz = a + b where a=2+53a = \sqrt[3]{2+\sqrt{5}} and b=2−53b = \sqrt[3]{2-\sqrt{5}}. Cubing both sides: z3=a3+b3+3ab(a+b)=(2+5)+(2−5)+3(2+5)(2−5)3z=4+34−53z=4−3zz^3 = a^3 + b^3 + 3ab(a+b) = (2+\sqrt{5}) + (2-\sqrt{5}) + 3\sqrt[3]{(2+\sqrt{5})(2-\sqrt{5})} z = 4 + 3\sqrt[3]{4-5} z = 4 - 3z. Solving z3+3z−4=0z^3 + 3z - 4 = 0 gives (z−1)(z2+z+4)=0(z-1)(z^2 + z + 4) = 0, whose unique real root is z=1z = 1.
4
Substitute all values into the given expression x3+y3x2+y2−z\frac{x^3 + y^3}{x^2 + y^2 - z}
\frac{52}{14 - 1} = \frac{52}{13} = 4
Direct numerical evaluation of the simplified algebraic components.

Key Concept

Simplification of Nested Square and Cube Surds using Algebraic Identities
Estimated Time:3m 0s
Question 33Question →

Three storage vessels in a chemical laboratory contain 1235\frac{12}{35} litres, 1825\frac{18}{25} litres, and 2449\frac{24}{49} litres of different reagents. What is the maximum capacity (in litres) of a single measuring container that can be used to measure the exact quantity of reagent from any of the three vessels without leaving any remainder?

Show answer & explanation

Answer: 61225\frac{6}{1225} litres

Answer

The maximum capacity of the measuring container is 61225\frac{6}{1225} litres.
To find the largest measuring unit that can measure fractional quantities exactly, we need the HCF of those fractions. The HCF of fractions is defined as HCF of numeratorsLCM of denominators\frac{\text{HCF of numerators}}{\text{LCM of denominators}}. Calculating HCF(12,18,24)=6\text{HCF}(12, 18, 24) = 6 and LCM(35,25,49)=1225\text{LCM}(35, 25, 49) = 1225 yields 61225\frac{6}{1225} litres.

Step-by-Step Solution

1
Identify the mathematical rule required for measuring fractional quantities exactly.
To find the maximum capacity of a container that measures given fractional quantities an exact number of times, compute the HCF of the fractions: HCF(ab,cd,ef)=HCF(a,c,e)LCM(b,d,f)\text{HCF}\left(\frac{a}{b}, \frac{c}{d}, \frac{e}{f}\right) = \frac{\text{HCF}(a, c, e)}{\text{LCM}(b, d, f)}.
The measuring container's capacity must be a common divisor of all three fractional volumes.
2
Calculate the HCF of the numerators 1212, 1818, and 2424.
HCF(12,18,24)=6\text{HCF}(12, 18, 24) = 6.
The prime factorizations are 12=22×312 = 2^2 \times 3, 18=2×3218 = 2 \times 3^2, and 24=23×324 = 2^3 \times 3. The highest common power of prime factors shared by all three numbers is 21×31=62^1 \times 3^1 = 6.
3
Calculate the LCM of the denominators 3535, 2525, and 4949.
LCM(35,25,49)=1225\text{LCM}(35, 25, 49) = 1225.
The prime factorizations are 35=5×735 = 5 \times 7, 25=5225 = 5^2, and 49=7249 = 7^2. The LCM takes the highest power of each prime: 52×72=25×49=12255^2 \times 7^2 = 25 \times 49 = 1225.
4
Substitute the calculated HCF of numerators and LCM of denominators into the fraction HCF formula.
HCF=61225\text{HCF} = \frac{6}{1225} litres.
Dividing the HCF of the numerators by the LCM of the denominators gives the exact maximum volume.

Key Concept

HCF of Fractions
Estimated Time:2m 0s
Question 34Question →
What is the final numerical value obtained by evaluating the following mathematical expression strictly according to the BODMAS rule?
50−[52÷{4.5+45×(3.7−1.8−0.6‾)}]50 - \left[ 52 \div \left\{ 4.5 + \frac{4}{5} \times \left( 3.7 - \overline{1.8 - 0.6} \right) \right\} \right]
Show answer & explanation

Answer: 42

Answer

42
Following VBODMAS rule strictly:
1. Vinculum: 1.8 - 0.6 = 1.2
2. Round brackets: 3.7 - 1.2 = 2.5
3. Multiplication in curly brackets: (4/5) * 2.5 = 2
4. Addition in curly brackets: 4.5 + 2 = 6.5
5. Square bracket division: 52 / 6.5 = 8
6. Outer subtraction: 50 - 8 = 42.

Step-by-Step Solution

1
Evaluate the vinculum (bar line above numbers)
\overline{1.8 - 0.6} = 1.2
According to BODMAS (or VBODMAS), the vinculum takes precedence over round brackets.
2
Simplify the terms inside the parentheses (round brackets)
3.7 - 1.2 = 2.5
Parentheses are evaluated next after the vinculum.
3
Evaluate the multiplication inside the curly brackets
\frac{4}{5} \times 2.5 = 0.8 \times 2.5 = 2.0
Within curly brackets, multiplication precedes addition.
4
Complete the addition inside the curly brackets
4.5 + 2.0 = 6.5
Completing all operations within the curly brackets.
5
Evaluate the division inside the square brackets
52÷6.5=852 \div 6.5 = 8
Simplifying the entire square bracket term.
6
Perform the final subtraction outside all brackets
50 - 8 = 42
Final arithmetic step to find the value of the expression.

Key Concept

Order of Operations (BODMAS / VBODMAS)
Question 35Question →
Calculate the exact numerical value of the following expression by applying the standard BODMAS/VBODMAS order of operations:
45% of 160−[12.5+{34×(36÷3−4.2+1.8‾)+5.5}÷0.5]45\% \text{ of } 160 - \left[ 12.5 + \left\{ \frac{3}{4} \times \left( 36 \div 3 - \overline{4.2 + 1.8} \right) + 5.5 \right\} \div 0.5 \right]
Show answer & explanation

Answer: 39.5

Answer

The simplified numerical value of the given expression is 39.5.
Evaluating step by step according to VBODMAS rules yields 39.5. First, the vinculum gives 6. The round bracket gives 36 ÷ 3 - 6 = 6. The curly bracket gives (3/4) × 6 + 5.5 = 10. The square bracket gives 12.5 + 10 ÷ 0.5 = 32.5. Finally, 45% of 160 = 72, and 72 - 32.5 = 39.5.

Step-by-Step Solution

1
Evaluate the expression under the vinculum (bar)
\overline{4.2 + 1.8} = 6
The vinculum acts as a top-priority bracket, so the addition under the bar must be executed before other operations.
2
Simplify the innermost round brackets ( )
36 \div 3 - 6 = 12 - 6 = 6
Inside the round bracket, division precedes subtraction according to BODMAS.
3
Simplify the curly brackets { }
\frac{3}{4} \times 6 + 5.5 = 4.5 + 5.5 = 10
Inside the curly bracket, multiplication of fractions precedes addition.
4
Simplify the square brackets [ ]
12.5 + 10 \div 0.5 = 12.5 + 20 = 32.5
Inside the square bracket, division by decimal (10 / 0.5 = 20) takes precedence over addition.
5
Calculate the percentage term ('of' operation)
45\% \text{ of } 160 = 0.45 \times 160 = 72
The 'of' operation represents multiplication for percentage evaluation.
6
Perform the final subtraction
72 - 32.5 = 39.5
Subtract the fully simplified bracket result from the percentage term.

Key Concept

BODMAS / VBODMAS Rule with Nested Brackets, Percentages, and Vinculum
Estimated Time:2m 0s
Question 36Question →
What is the simplified numerical value of the mathematical expression 36−[18÷{6−(4−3−1‾)}×2]36 - \left[ 18 \div \left\{ 6 - \left( 4 - \overline{3 - 1} \right) \right\} \times 2 \right] when evaluated strictly adhering to the VBODMAS rule?
Show answer & explanation

Answer: 2727

Answer

The simplified numerical value of the given expression is 27.
First resolve the vinculum 3−1‾=2\overline{3-1} = 2. Next, simplify the round bracket (4−2)=2(4-2) = 2, then the curly bracket {6−2}=4\{6-2\} = 4. Inside the square bracket, evaluate division before multiplication from left to right: 18÷4=4.518 \div 4 = 4.5, and 4.5×2=94.5 \times 2 = 9. Finally, subtract from 36 to get 36−9=2736 - 9 = 27.

Step-by-Step Solution

1
Evaluate the expression under the bar/vinculum
\overline{3 - 1} = 2
According to VBODMAS rules, vinculum (bar) has highest priority over standard operations.
2
Simplify the innermost round bracket
(4 - 2) = 2
Resolve terms within parenthesis after vinculum.
3
Simplify the curly bracket
\{6 - 2\} = 4
Perform subtraction inside curly braces.
4
Evaluate operations inside the square bracket from left to right
[18 \div 4 \times 2] = 4.5 \times 2 = 9
Division and multiplication have equal priority and must be evaluated left to right.
5
Perform the final outer subtraction
36 - 9 = 27
Complete the remaining operation.

Key Concept

VBODMAS / BODMAS Order of Operations
Question 37Question →
What is the numerical value of the mathematical expression 80−[15+(12÷3×2)]80 - [15 + (12 \div 3 \times 2)] when simplified using the standard BODMAS rule?
Show answer & explanation

Answer: 57

Answer

The simplified numerical value of the given expression is 57.
Evaluating the expression following the strict BODMAS order of operations: first simplify the innermost round brackets by doing division (12÷3=412 \div 3 = 4) then multiplication (4×2=84 \times 2 = 8). Next, add the numbers inside the square brackets (15+8=2315 + 8 = 23). Finally, subtract 23 from 80 to obtain 57.

Step-by-Step Solution

1
Perform division inside the round brackets.
12÷3=412 \div 3 = 4
According to the BODMAS rule, division takes precedence over multiplication within brackets.
2
Perform multiplication to complete the round bracket evaluation.
4×2=84 \times 2 = 8
Complete the operations contained within the innermost parentheses.
3
Perform addition inside the square brackets.
15+8=2315 + 8 = 23
Simplify the terms remaining inside the square brackets.
4
Subtract the result of the brackets from the outer value.
80−23=5780 - 23 = 57
Perform final subtraction to determine the simplified answer.

Key Concept

BODMAS Rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction)
Question 38Question →

Three automated robotic arms in an assembly line complete their respective sorting cycles in 1415\frac{14}{15} minutes, 2120\frac{21}{20} minutes, and 4225\frac{42}{25} minutes. If all three robotic arms start a sorting cycle simultaneously at 9:00:00 AM, after what minimum time interval will all three arms simultaneously begin a new sorting cycle again?

Show answer & explanation

Answer: 88 minutes 2424 seconds

Answer

88 minutes 2424 seconds
To find the time when events with fractional period lengths coincide again, we compute the LCM of the fraction time intervals. The formula for the LCM of fractions is LCM of numeratorsHCF of denominators\frac{\text{LCM of numerators}}{\text{HCF of denominators}}. Finding LCM(14,21,42)=42\text{LCM}(14, 21, 42) = 42 and HCF(15,20,25)=5\text{HCF}(15, 20, 25) = 5 yields 425=8.4\frac{42}{5} = 8.4 minutes, which equals 88 minutes 2424 seconds.

Step-by-Step Solution

1
Identify the mathematical operation required for simultaneous recurrence
The required minimum time interval is the Least Common Multiple (LCM) of the three cycle times: LCM(1415,2120,4225)\text{LCM}\left(\frac{14}{15}, \frac{21}{20}, \frac{42}{25}\right) minutes.
Simultaneous future events occurring at periodic intervals require finding the LCM of the given time periods.
2
Apply the formula for the LCM of fractions
LCM(ab,cd,ef)=LCM(a,c,e)HCF(b,d,f)\text{LCM}\left(\frac{a}{b}, \frac{c}{d}, \frac{e}{f}\right) = \frac{\text{LCM}(a, c, e)}{\text{HCF}(b, d, f)}
For simplified fractions, the LCM is determined by dividing the LCM of all numerators by the HCF of all denominators.
3
Calculate the LCM of the numerators (14,21,42)(14, 21, 42)
14=2×714 = 2 \times 7, 21=3×721 = 3 \times 7, 42=2×3×742 = 2 \times 3 \times 7. Thus, LCM(14,21,42)=2×3×7=42\text{LCM}(14, 21, 42) = 2 \times 3 \times 7 = 42.
The smallest positive integer divisible by 14, 21, and 42 is 42.
4
Calculate the HCF of the denominators (15,20,25)(15, 20, 25)
15=3×515 = 3 \times 5, 20=22×520 = 2^2 \times 5, 25=5225 = 5^2. Thus, HCF(15,20,25)=5\text{HCF}(15, 20, 25) = 5.
The greatest integer that divides 15, 20, and 25 without remainder is 5.
5
Combine the results and convert to minutes and seconds
LCM=425=8.4\text{LCM} = \frac{42}{5} = 8.4 minutes. Converting 0.40.4 minutes to seconds gives 0.4×60=240.4 \times 60 = 24 seconds. Total time = 88 minutes 2424 seconds.
Multiplying the fractional minute by 60 converts it into exact seconds.

Key Concept

LCM of Fractions
Question 39Question →

If x=7+37−3x = \frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}} and y=7−37+3y = \frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}, what is the value of x2+y2x^2 + y^2?

Show answer & explanation

Answer: 23

Answer

The value of x2+y2x^2 + y^2 is 23.
By rationalizing the denominators of xx and yy, we obtain x=5+212x = \frac{5+\sqrt{21}}{2} and y=5−212y = \frac{5-\sqrt{21}}{2}. Adding these gives x+y=5x+y=5, and multiplying them gives xy=1xy=1. Using the identity x2+y2=(x+y)2−2xyx^2+y^2 = (x+y)^2 - 2xy, we calculate 52−2(1)=235^2 - 2(1) = 23.

Step-by-Step Solution

1
Rationalize the expressions for xx and yy by multiplying the numerator and denominator by their respective conjugates.
x=(7+3)27−3=10+2214=5+212x = \frac{(\sqrt{7}+\sqrt{3})^2}{7-3} = \frac{10 + 2\sqrt{21}}{4} = \frac{5 + \sqrt{21}}{2} and y=(7−3)27−3=10−2214=5−212y = \frac{(\sqrt{7}-\sqrt{3})^2}{7-3} = \frac{10 - 2\sqrt{21}}{4} = \frac{5 - \sqrt{21}}{2}
Eliminating surds from the denominator simplifies addition and multiplication.
2
Calculate the sum (x+y)(x + y) and the product (x⋅y)(x \cdot y) of xx and yy.
x+y=5+212+5−212=5x + y = \frac{5 + \sqrt{21}}{2} + \frac{5 - \sqrt{21}}{2} = 5 and x⋅y=(5+21)(5−21)4=25−214=1x \cdot y = \frac{(5 + \sqrt{21})(5 - \sqrt{21})}{4} = \frac{25 - 21}{4} = 1
Simplifying the symmetric expressions reduces computation complexity.
3
Apply the algebraic identity x2+y2=(x+y)2−2xyx^2 + y^2 = (x + y)^2 - 2xy.
x2+y2=(5)2−2(1)=25−2=23x^2 + y^2 = (5)^2 - 2(1) = 25 - 2 = 23
Using the identity avoids evaluating complex squares directly.

Key Concept

Rationalization of surds and application of symmetric algebraic identities
Question 40Question →

If 2x−1+2x+1=3202^{x-1} + 2^{x+1} = 320, what is the value of xx?

Show answer & explanation

Answer: 77

Answer

77
Factoring 2x−12^{x-1} from the left side gives 2x−1(1+22)=5⋅2x−1=3202^{x-1}(1 + 2^2) = 5 \cdot 2^{x-1} = 320. Dividing both sides by 55 yields 2x−1=64=262^{x-1} = 64 = 2^6. Equating the powers gives x−1=6x - 1 = 6, which solves to x=7x = 7.

Step-by-Step Solution

1
Factor out the common exponential term from the left-hand side
2x−1(1+22)=320  ⟹  2x−1(1+4)=320  ⟹  5⋅2x−1=3202^{x-1}(1 + 2^2) = 320 \implies 2^{x-1}(1 + 4) = 320 \implies 5 \cdot 2^{x-1} = 320
Applying the distributive law of exponents to simplify the sum.
2
Divide both sides by 5
2x−1=3205=642^{x-1} = \frac{320}{5} = 64
Isolate the exponential term with base 2.
3
Express 64 as a power of 2 and equate the exponents
64=26  ⟹  x−1=6  ⟹  x=764 = 2^6 \implies x - 1 = 6 \implies x = 7
Since the bases are equal (22), the exponents must be equal.

Key Concept

Laws of Indices and Exponential Equations
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