SSC CGL Tier 1 2011 Quantitative Aptitude Questions with Solutions

Explanation

The expression inside the cube root can be expanded by distributing p as follows: p(p² + 3p + 3) + 1 = p³ + 3p² + 3p + 1

Using the algebraic identity for the cube of a binomial, (a + b)³ = a³ + 3a²b + 3ab² + b³, we can identify that: p³ + 3p²(1) + 3p(1²) + 1³ = (p + 1)³

Therefore, the entire expression simplifies to: ³√((p + 1)³) = p + 1

Given that p = 124, substituting the value gives: 124 + 1 = 125

Explanation

Let the two natural numbers be x and y. The sum S is expressed as S = 18x + 21y. By factoring out the greatest common divisor of 18 and 21, which is 3, the expression becomes S = 3(6x + 7y). This shows that the sum must be a multiple of 3. According to the divisibility rule for 3, a number is divisible by 3 if the sum of its digits is divisible by 3. For the number 2007, the sum of the digits is 2 + 0 + 0 + 7 = 9. Since 9 is divisible by 3, the number 2007 is a multiple of 3. For the equation 6x + 7y = 669, natural number solutions exist, such as x = 108 and y = 3, which satisfy the condition.

Explanation

Let the two numbers be x and y. We are given that their product xy = 45 and their difference x - y = 4.

To find the sum of squares x² + y², we use the algebraic identity for the square of a difference: (x - y)² = x² + y² - 2xy

By rearranging the terms to isolate the sum of squares, we get: x² + y² = (x - y)² + 2xy

Substituting the given values into the equation: x² + y² = (4)² + 2(45) x² + y² = 16 + 90 x² + y² = 106

Explanation

The expression is a nested radical, which is solved by simplifying from the innermost square root outwards.

Step 1: Evaluate the innermost square root, √(169) = 13. Step 2: Substitute this value into the next layer, √(108 + 13) = √(121) = 11. Step 3: Continue to the next outer layer, √(38 + 11) = √(49) = 7. Step 4: Evaluate the next layer, √(57 + 7) = √(64) = 8. Step 5: Solve the final square root, √(8 + 8) = √(16) = 4.

Explanation

To find the square root of an expression of the form a + √b, we use the algebraic identity (x + y)² = x² + y² + 2xy. We represent the given expression 14 + 6√5 as a perfect square by identifying x² + y² = 14 and 2xy = 6√5. From the second equation, xy = 3√5. We test the values x = 3 and y = √5. Calculating the sum of their squares gives x² + y² = 3² + (√5)² = 9 + 5 = 14. This matches the rational term in the expression. Consequently, 14 + 6√5 can be rewritten as (3 + √5)². Taking the square root of (3 + √5)² yields 3 + √5.

Explanation

To calculate the remainder of 2³¹ divided by 5, we use the property of power cyclicity or Euler's Totient Theorem. Since 5 is a prime number, the totient φ(5) is 5 - 1 = 4, which implies that 2⁴ ≡ 1 5 because 2 and 5 are coprime. We can rewrite the power by dividing the exponent 31 by this cycle length of 4, giving 31 = (4 × 7) + 3. The expression thus becomes 2³¹ = (2⁴)⁷ × 2³. Substituting the congruency results in 1⁷ × 2³ 5, which simplifies to 1 × 8 = 8. Dividing 8 by 5 leaves a remainder of 3. Alternatively, observing the pattern of remainders for 2ⁿ 5 shows a repeating sequence of 2, 4, 3, and 1. Since 31 4 = 3, the remainder corresponds to the third value in this sequence, which is 3.

Explanation

The unit digit of any expression in the form aⁿ depends only on the unit digit of the base a. For the expression 122¹⁷³, the unit digit of the base is 2. The unit digits of the powers of 2 follow a repeating cycle of four: 2¹ = 2, 2² = 4, 2³ = 8, and 2⁴ = 16 (unit digit 6). This cycle (2, 4, 8, 6) repeats every four powers. To determine the unit digit of 122¹⁷³, we divide the exponent 173 by the cycle length 4. The calculation 173 = (43 × 4) + 1 shows a remainder of 1. A remainder of 1 corresponds to the first position in the cycle, which is 2.

Explanation

The expression consists of two parts that can be simplified using standard algebraic identities and rationalization. For the first two terms, we apply the identity (a+b)/(a-b) + (a-b)/(a+b) = 2(a² + b²){a² - b²}. Substituting a = 2 and b = √3:

2 + √3{2 - √3} + 2 - √3{2 + √3} = 2(2² + (√3)²){2² - (√3)²} = (2(4 + 3))/(4 - 3) = ¹⁴⁄₁ = 14

For the third term, we rationalize the denominator by multiplying both the numerator and the denominator by the conjugate (√3 + 1):

√3 + 1{√3 - 1} × √3 + 1{√3 + 1} = (√3 + 1)²{(√3)² - 1²} = 3 + 1 + 2√3{3 - 1} = 4 + 2√3{2} = 2 + √3

Combining the results of both parts, we get:

14 + (2 + √3) = 16 + √3

Explanation

The operation is defined as a b = 2a + 3b - ab. To find the value of (3 5 + 5^* 3), evaluate each term separately based on the given formula.

For the first term 3^* 5, substitute a = 3 and b = 5: 3^* 5 = 2(3) + 3(5) - (3 × 5) 3^* 5 = 6 + 15 - 15 = 6

For the second term 5^* 3, substitute a = 5 and b = 3: 5^* 3 = 2(5) + 3(3) - (5 × 3) 5^* 3 = 10 + 9 - 15 = 4

Summing the results of the two operations: 3 5 + 5 3 = 6 + 4 = 10

Explanation

The given expression follows the algebraic identity for the sum of two cubes: a³ + b³ = (a + b)(a² - ab + b²). This can be rearranged to the form:

(a³ + b³)/(a² - ab + b²) = a + b

In the provided simplification problem, let a = 0.0347 and b = 0.9653. The numerator represents a³ + b³ and the denominator represents the quadratic factor a² - ab + b². Substituting these values into the identity, the entire expression simplifies directly to the sum of a and b:

0.0347 + 0.9653 = 1.0000

By identifying the structure of the expression as a standard algebraic form, the complex decimal multiplications and divisions are bypassed in favor of a simple addition.

Explanation

The area of an equilateral triangle is given by the formula A = √3{4} s², where s is the side of the triangle. Substituting the given area, we have √3{4} s² = 121√3. Dividing both sides by √3 and multiplying by 4 yields s² = 484, which gives a side length of s = 22 cm. The total length of the wire is equal to the perimeter of this triangle, which is 3s = 3 × 22 = 66 cm.

When the same wire is bent into the form of a circle, the circumference 2π r must equal the length of the wire. Using π = ²²⁄₇, we set up the equation 2 × ²²⁄₇ × r = 66. Simplifying this, we find ⁴⁴⁄₇ r = 66, which results in r = (66 × 7)/(44) = (3 × 7)/(2) = 10.5 cm. The area of the circle is then calculated using the formula A = π r². Substituting the values, we get A = ²²⁄₇ × ²¹⁄₂ × ²¹⁄₂ = (11 × 3 × 21)/(2) = ⁶⁹³⁄₂ = 346.5 cm2.

Explanation

When a geometric solid is reshaped into another, the total volume remains invariant. The volume of the cone is calculated using the formula V = ⅓ π r² h, and the volume of the resulting sphere is V = ⁴⁄₃ π R³. Given the cone has a radius r = 6 cm and a height h = 24 cm, we equate the two volume expressions:

⅓ π (6)² (24) = ⁴⁄₃ π R³

The constants ⅓ and π are common to both sides and can be canceled, simplifying the equation to:

36 × 24 = 4 R³

Dividing both sides by 4:

R³ = 36 × 6 R³ = 216

Taking the cube root of 216 yields:

R = ³√(216) = 6 cm

Explanation

Let the two numbers be 13x and 13y, where x and y are co-prime integers. This ensures that their Highest Common Factor (H.C.F.) remains 13. According to the problem, the product of these two numbers is 2028:

13x × 13y = 2028 169xy = 2028 xy = ²⁰²⁸⁄₁₆₉ = 12

To find the number of pairs, we identify the pairs of factors of 12 that are co-prime to each other. The possible factor pairs for 12 are:

(1, 12): Since HCF(1, 12) = 1, these are co-prime. (2, 6): Since HCF(2, 6) = 2, these are not co-prime. (3, 4): Since HCF(3, 4) = 1, these are co-prime.

The valid pairs for (x, y) are (1, 12) and (3, 4). Consequently, there are 2 such pairs of numbers that satisfy the given conditions.

Explanation

Since the two vessels have equal volume, let the capacity of each vessel be the Least Common Multiple (LCM) of the sum of the ratio parts. For the first vessel, the sum is 3 + 4 = 7. For the second vessel, the sum is 5 + 3 = 8. The LCM of 7 and 8 is 56. Thus, assume each vessel contains 56 units of the mixture.

In the first vessel, the quantity of water is ³⁄₇ × 56 = 24 units and the quantity of milk is ⁴⁄₇ × 56 = 32 units. In the second vessel, the quantity of water is ⅝ × 56 = 35 units and the quantity of milk is ⅜ × 56 = 21 units.

When these are poured into a third vessel, the total quantity of water becomes 24 + 35 = 59 units and the total quantity of milk becomes 32 + 21 = 53 units. Therefore, the ratio of water to milk in the third vessel is 59:53.

Explanation

Let the current age of the son be x years. Since the father is three times as old as the son, the father's current age is 3x years.

In 15 years, the son's age will be x + 15 and the father's age will be 3x + 15. According to the problem, the father will then be twice as old as the son: 3x + 15 = 2(x + 15) 3x + 15 = 2x + 30 x = 15

The current age of the son is 15 years, and the current age of the father is 3 × 15 = 45 years. The sum of their present ages is 15 + 45 = 60 years.

Explanation

The time interval after which all three bells will ring together again is determined by the Least Common Multiple (LCM) of their individual ringing intervals. Given the intervals are 20, 30, and 40 minutes, we calculate their LCM using prime factorization:

20 = 2² × 5 30 = 2 × 3 × 5 40 = 2³ × 5

The LCM(20, 30, 40) = 2³ × 3 × 5 = 8 × 3 × 5 = 120 minutes.

Since there are 60 minutes in an hour, 120 minutes is equivalent to 120 / 60 = 2 hours. The bells first rang together at 11 a.m., so the next time they will ring together is 2 hours later, which is 1 p.m.

Explanation

Let the total work be 60 units, which is the LCM of 12 and 15. The combined efficiency of A and B is 60 / 12 = 5 units per day, and the combined efficiency of B and C is 60 / 15 = 4 units per day. Subtracting the second efficiency from the first gives (A + B) - (B + C) = 5 - 4, which simplifies to A - C = 1 unit per day.

Given that A's efficiency is twice that of C (A = 2C), we substitute this into the equation: 2C - C = 1, which means C = 1 unit per day. To find B's efficiency, we use the equation for B and C: B + 1 = 4, yielding B = 3 units per day. The time required for B alone to finish the work is the total work divided by B's efficiency, which is 60 / 3 = 20 days.

Explanation

Let the total work be the least common multiple of the given time periods, which is LCM(12, 15, 20) = 60 units. The daily work efficiency of each pair is calculated by dividing the total work by the time taken. The efficiency of A and B is 60 / 12 = 5 units per day, B and C is 60 / 15 = 4 units per day, and C and A is 60 / 20 = 3 units per day. Adding these efficiencies gives 2(A + B + C) = 5 + 4 + 3 = 12 units per day, which implies the combined efficiency of A, B, and C is 12 / 2 = 6 units per day. The time taken by A, B, and C to complete the work together is determined by dividing the total work by their combined efficiency: 60 / 6 = 10 days.

Explanation

Let the efficiency of B be 2 units/day. Since A is 50% as efficient as B, the efficiency of A is 1 unit/day. According to the problem, C does half the work done by A and B together, so the efficiency of C is (1 + 2)/(2) = 1.5 units/day.

The total work is calculated by multiplying the efficiency of C by the time it takes to complete the work alone. Total Work = 1.5 × 20 = 30 units. The combined efficiency of A, B, and C working together is 1 + 2 + 1.5 = 4.5 units/day.

The time taken by A, B, and C together to complete the work is (30)/(4.5). Multiplying the numerator and denominator by 10 gives ³⁰⁰⁄₄₅. Simplifying this fraction by dividing both parts by 15 results in ²⁰⁄₃ days.

Explanation

The initial ratio of water to glycerine is 1:3, making a total of 1 + 3 = 4 parts. Given the total volume is 240 cc, the volume of one part is calculated as 240 / 4 = 60 cc. This means the mixture initially contains 60 cc of water and 180 cc of glycerine.

In the new ratio of 2:3, the amount of glycerine remains unchanged because only water is added. Since the glycerine still accounts for 3 parts in the new ratio, and its volume is 180 cc, one part in the new ratio is 180 / 3 = 60 cc. The new volume of water is represented by 2 parts, which equals 2 × 60 = 120 cc. The quantity of water added is the difference between the new and old volumes: 120 cc - 60 cc = 60 cc.

Alternatively, since the glycerine parts remain constant at 3, the increase in water is simply from 1 part to 2 parts, which is an increase of 1 part. As previously determined, 1 part equals 60 cc, so the added water is 60 cc.

Explanation

Let the present ages of Maya and Chhaya be 6x and 5x respectively. According to the problem, fifteen years from now, the ratio of their ages will be 9:8. We can set up the following equation based on this future ratio:

(6x + 15)/(5x + 15) = ⁹⁄₈

Cross-multiplying the terms gives:

8(6x + 15) = 9(5x + 15) 48x + 120 = 45x + 135

Subtracting 45x and 120 from both sides:

3x = 15 x = 5

Maya's present age is 6x. Substituting the value of x:

Maya's age = 6 × 5 = 30 years

Alternatively, using the ratio unit method, the increase in ratio units for both Maya (9 - 6 = 3) and Chhaya (8 - 5 = 3) is the same. This increase of 3 units corresponds to the 15 years that have passed. Therefore, 1 unit equals 5 years. Since Maya's present age is 6 units, her age is 6 × 5 = 30 years.

Explanation

The ratio of income to expenditure is given as 10:7. Since Income = Expenditure + Savings, the savings part of the ratio is 10 - 7 = 3 units. The expenditure is given as Rs. 10,500, which corresponds to 7 units in the ratio. Therefore, the value of one unit is 10,500 / 7 = 1,500. To find the total savings, we multiply the savings units by the value per unit: 3 × 1,500 = 4,500. This confirms the savings of the family are Rs. 4,500.

Explanation

Using the rule of alligation, the ratio of the number of students in Section A to the number of students in Section B is determined by the absolute differences between the individual section averages and the combined average of the two sections.

Ratio n_A : n_B = (Average_Cₒₘ_bᵢₙₑ_d - Average_B) : (Average_A - Average_Cₒₘ_bᵢₙₑ_d)

Substituting the given values: Ratio n_A : n_B = (74 - 70) : (77.5 - 74) Ratio n_A : n_B = 4 : 3.5

To convert this to a ratio of integers, multiply both terms by 2: n_A : n_B = 8 : 7

Alternatively, this can be solved using the weighted average formula where n_A and n_B represent the number of students in each section: Average_Cₒₘ_bᵢₙₑ_d = (n_A × Average_A + n_B × Average_B)/(n_A + n_B) 74 = (77.5 n_A + 70 n_B)/(n_A + n_B) 74 (n_A + n_B) = 77.5 n_A + 70 n_B 74 n_A + 74 n_B = 77.5 n_A + 70 n_B 74 n_B - 70 n_B = 77.5 n_A - 74 n_A 4 n_B = 3.5 n_A (n_A)/(n_B) = (4)/(3.5) = ⁴⁰⁄₃₅ = ⁸⁄₇

Explanation

The change in the total weight is calculated by finding the difference between the correct value and the misread value, which is 87 - 78 = 9 kg. To find the correction for the average, this difference is divided by the total number of boys in the group: ⁹⁄₂₀ = 0.45 kg. The correct average weight is obtained by adding this increment to the initial average, resulting in 89.4 + 0.45 = 89.85 kg.

Explanation

The distance covered by a wheel in one complete revolution is equal to its circumference C. The circumference of a circle is given by the formula C = π d, where d is the diameter. Using d = 98 cm and π = ²²⁄₇, the circumference is:

C = ²²⁄₇ × 98 = 22 × 14 = 308 cm

The total distance to be covered is 1540 m. Converting this distance into centimeters to match the units of the circumference, we get:

1540 m = 1540 × 100 = 154000 cm

The number of revolutions n required to cover the total distance D is calculated by the formula n = (D)/(C):

n = ¹⁵⁴⁰⁰⁰⁄₃₀₈ = 500

Explanation

In an equilateral triangle ABC, all sides are equal to 10 cm and all internal angles are 60 degrees. Since side BC is trisected at D, the length BD is one-third of BC, which gives BD = 10/3 cm.

Applying the Law of Cosines in triangle ABD: AD² = AB² + BD² - 2(AB)(BD) cos B

Substituting the known values AB = 10, BD = 10/3, and ∠ B = 60°: AD² = 10² + (10/3)² - 2(10)(10/3) cos 60° AD² = 100 + 100/9 - 2(100/3)(1/2) AD² = 100 + 100/9 - 100/3

Taking a common denominator of 9: AD² = (900 + 100 - 300) / 9 AD² = 700 / 9

Taking the square root: AD = √(700/9) = 10√7 / 3

Explanation

The relation between the marked price (MP) and the cost price (CP) when a discount and gain are given is defined by the formula:

(MP)/(CP) = (100 + Gain%)/(100 - Discount%)

Substituting the given values, where CP = Rs. 800, Gain% = 12.5, and Discount% = 10:

(MP)/(800) = (100 + 12.5)/(100 - 10)

(MP)/(800) = (112.5)/(90)

MP = ¹¹²⁵⁄₉₀₀ × 800

MP = ⁵⁄₄ × 800 = 5 × 200

MP = 1000

Alternatively, the selling price (SP) can be calculated first. At a gain of 12.5%, SP = 800 × (1 + 12.5/100) = 800 × 1.125 = 900. Since this SP is achieved after a 10% discount on the marked price, 0.9 × MP = 900, which gives MP = 1000.

Explanation

The list price of the article is Rs. 1500. After the first discount of 20%, the reduced price is calculated as: Price after first discount = 1500 × (1 - ²⁰⁄₁₀₀) = 1500 × 0.8 = Rs. 1200

To reach the final net price of Rs. 1104, an additional discount must be applied to this Rs. 1200. The amount of the additional discount required is: Additional discount amount = 1200 - 1104 = Rs. 96

The rate of this additional discount is determined by dividing the discount amount by the price after the first discount: Additional discount percentage = (⁹⁶⁄₁₂₀₀) × 100 = ⁹⁶⁄₁₂ = 8%

Explanation

Given the ratio (a)/(b) = (c)/(d) = (e)/(f) = k = 3 , we can express the numerator terms as a = 3b , c = 3d , and e = 3f . Squaring these relations gives a² = 9b² , c² = 9d² , and e² = 9f² . Substituting these squared values into the target expression:

(2a² + 3c² + 4e²)/(2b² + 3d² + 4f²) = (2(9b²) + 3(9d²) + 4(9f²))/(2b² + 3d² + 4f²)

By factoring out 9 from each term in the numerator, the expression simplifies to:

(9(2b² + 3d² + 4f²))/(2b² + 3d² + 4f²) = 9

This result is consistent with the general algebraic property where if (a)/(b) = (c)/(d) = (e)/(f) = k , then any expression of the form (xaⁿ + ycⁿ + zeⁿ)/(xbⁿ + ydⁿ + zfⁿ) will equal kⁿ . In this specific problem, k = 3 and n = 2 , which directly yields 3² = 9 .

Explanation

The total area to be painted consists of the four walls and the ceiling of the room. The lateral surface area representing the four walls is calculated using the formula 2(l + w) × h, where l is the length, w is the width, and h is the height. Substituting the given dimensions l = 4m, w = 3m, and h = 3m, the area of the walls is 2(4 + 3) × 3 = 6 × 7 = 42m². The area of the ceiling is the same as the area of the floor, calculated as l × w = 4 × 3 = 12m². By summing these values, the total area to be painted is 42m² + 12m² = 54m².

Explanation

Revenue is the product of the price of an article and the quantity sold (R = P × S). When two factors are multiplied, the net percentage change in their product is given by the formula x + y + ˣʸ⁄₁₀₀, where x and y are the percentage changes in the factors.

In this case, the price change x = -20% (reduction) and the sale change y = +80% (increase). Substituting these values into the formula:

Net effect = -20 + 80 + ((-20 × 80))/(100)

Net effect = 60 - ¹⁶⁰⁰⁄₁₀₀

Net effect = 60 - 16 = 44%

Alternatively, using the multiplier method, if the initial price and sale are both 100, the initial revenue is 100 × 100 = 10,000. After the changes, the new price is 80 and the new sale is 180. The new revenue becomes 80 × 180 = 14,400. The increase in revenue is 14,400 - 10,000 = 4,400, which is a 44% increase over the original 10,000.

Explanation

When expenditure remains constant, price and consumption are inversely proportional (E = P × C). If the price of a commodity increases by r%, the percentage reduction in consumption required to keep the total expenditure unchanged is given by the formula (r)/(100 + r) × 100%.

Substituting the given value r = 20:

Reduction in consumption = (20)/(100 + 20) × 100 Reduction in consumption = ²⁰⁄₁₂₀ × 100 Reduction in consumption = ⅙ × 100 = ⁵⁰⁄₃%

Alternatively, assume the initial price is 100 and initial consumption is 100 units, making the total expenditure 10,000. After a 20% increase, the new price is 120. To keep the expenditure at 10,000, the new consumption must be (10,000)/(120) = ²⁵⁰⁄₃ units. The reduction in consumption is 100 - ²⁵⁰⁄₃ = (300 - 250)/(3) = ⁵⁰⁄₃%.

Explanation

Let the total number of workers in the factory be represented by x. The problem states that 60% of these workers are above 30 years of age. Out of this specific group, 75% are males. Given that the total number of male workers above 30 years is 1350, we can set up the following linear equation:

x × ⁶⁰⁄₁₀₀ × ⁷⁵⁄₁₀₀ = 1350

To find x, simplify the percentages into fractions: x × ⅗ × ¾ = 1350 x × ⁹⁄₂₀ = 1350

Now, isolate x by multiplying both sides by the reciprocal of the fraction: x = (1350 × 20)/(9) x = 150 × 20 x = 3000

Explanation

Since distance is constant, speed and time are inversely proportional (S ∝ 1/T). When the speed is 3/4 of the usual speed, the time taken becomes the reciprocal, which is 4/3 of the usual time.

Let the usual time be T. The problem states that the man is 1 ½ hours late, which means the difference between the new time and the usual time is 1.5 hours.

⁴⁄₃T - T = 1.5

⅓T = 1.5

T = 1.5 × 3 = 4.5

The usual time 4.5 hours is equivalent to 4 ½ hours.

Explanation

Let the cost price (CP) of one orange be x. According to the question, the selling price (SP) of 10 oranges is equal to the cost price of 13 oranges, which is 13x. To find the profit or loss, we compare the prices for the same quantity of goods. The cost price of 10 oranges is 10x.

Profit is calculated as the difference between the selling price and the cost price: Profit = SP of 10 oranges - CP of 10 oranges = 13x - 10x = 3x.

The profit percentage is calculated using the formula: Profit % = (Profit)/(CP) × 100 Profit % = ³ˣ⁄₁₀ₓ × 100 = ³⁄₁₀ × 100 = 30%.

Alternatively, when the selling price of n items equals the cost price of m items, the profit percentage is given by (m - n)/(n) × 100. Substituting m = 13 and n = 10 yields (13 - 10)/(10) × 100 = 30%.

Explanation

The relationship between Cost Price (CP) and Marked Price (MP) is given by the formula (CP)/(MP) = (100 - D%)/(100 + P%), where D is the discount percentage and P is the profit percentage. Substituting the given values, we get (CP)/(MP) = (100 - 10)/(100 + 8) = ⁹⁰⁄₁₀₈. Simplifying this ratio by dividing both numerator and denominator by 18, we find (CP)/(MP) = ⅚.

If no discount is allowed, the item is sold at its Marked Price, making the new Selling Price (SP) equal to the MP. Using the ratio, we can consider the CP as 5 units and the new SP as 6 units. The profit made in this scenario is SP - CP = 6 - 5 = 1 unit. The gain percentage is then calculated as ((Profit)/(CP)) × 100 = (⅕) × 100 = 20%. The absolute value of the marked price is not necessary to determine the percentage change.

Explanation

The selling price (SP) of 20 apples is Rs. 1000, which means the SP per apple is 1000 / 20 = 50. Since there is a gain of 20%, the cost price (CP) per apple is calculated using the formula CP = (SP)/(1 + gain){100}. Substituting the values gives CP = (50)/(1.2) = ⁵⁰⁰⁄₁₂ per apple. To find the number of apples bought for Rs. 1000, divide the total amount by the cost price per apple: 1000 / ⁵⁰⁰⁄₁₂ = (1000 × 12)/(500) = 2 × 12 = 24.

A faster shortcut for problems where the total transaction amount (Rs. 1000) is constant is to use the inverse relationship between price and quantity. If the gain is 20%, the ratio of the selling price to the cost price per unit is 120:100 or 6:5. Consequently, the ratio of the quantity bought to the quantity sold must be 6:5 to maintain the same total cost. Let the quantity bought be x. Then x / 20 = 6 / 5, which solves to x = 24.

Explanation

When squares of side x = 4 cm are cut from the corners of a rectangular sheet of length L = 40 cm and width W = 15 cm, the dimensions of the resulting open box are determined by subtracting twice the side of the square from the original dimensions. The new length l becomes 40 - 2(4) = 32 cm, and the new width w becomes 15 - 2(4) = 7 cm. The height h of the box corresponds to the side of the square removed, which is 4 cm. The volume V of the rectangular box is calculated using the formula V = l × w × h. Substituting the values, we get V = 32 cm × 7 cm × 4 cm = 896 cm³.

Explanation

The given ratio is 1 : ⅓ : ⅙. To convert this into a simpler whole-number ratio, multiply each term by the least common multiple of the denominators, which is 6. This calculation yields 1 × 6 : ⅓ × 6 : ⅙ × 6, resulting in the ratio 6 : 2 : 1.

The total sum of the ratio parts is 6 + 2 + 1 = 9. The entire quantity of 78 is divided into these 9 parts. The value of a single part is calculated as ⁷⁸⁄₉, which simplifies to ²⁶⁄₃.

The middle part corresponds to the second term in the ratio, which is 2. Therefore, the value of the middle part is 2 × ²⁶⁄₃ = ⁵²⁄₃. Converting this improper fraction into a mixed number gives 17 ⅓.

Explanation

The formula for Simple Interest is SI = (P × R × T)/(100), where P is the principal, R is the rate of interest per annum, and T is the time in years. According to the problem, the simple interest is one-ninth of the principal, so SI = (P)/(9). Additionally, the number of years is equal to the rate per cent per annum, so T = R.

Substituting these values into the standard formula: (P)/(9) = (P × R × R)/(100)

By dividing both sides by P, the equation simplifies to: ⅑ = (R²)/(100)

Rearranging the terms to solve for R²: R² = ¹⁰⁰⁄₉

Taking the square root of both sides to find the value of R: R = √(¹⁰⁰⁄₉) = ¹⁰⁄₃

Thus, the rate of interest per annum is ¹⁰⁄₃%.

Explanation

For a period of 2 years, the difference between compound interest and simple interest is given by the formula D = P × ((R)/(100))², where D is the difference, P is the principal sum, and R is the annual rate of interest.

Given that D = 48 and R = 20, we substitute these values into the formula: 48 = P × (²⁰⁄₁₀₀)²

Simplifying the fraction: 48 = P × (⅕)² 48 = P × ¹⁄₂₅

Solving for P: P = 48 × 25 = 1200

Using the effective interest rate method, the simple interest for 2 years at 20% is 20% × 2 = 40%. The effective compound interest for 2 years is calculated as x + y + ˣʸ⁄₁₀₀, which gives 20 + 20 + (20 × 20)/(100) = 44%. The difference between the two rates is 44% - 40% = 4%. Since 4% of the sum equals Rs. 48, the total sum is ⁴⁸⁄₄ × 100 = 1200.

Explanation

The distance D traveled can be calculated using the relationship between speed, distance, and time difference. When an object travels at two different speeds v₁ and v₂ resulting in a total time difference Δ t, the distance is given by the formula D = v₁ × v₂{|v₁ - v₂|} × Δ t.

In this problem, the two speeds are v₁ = 30 km/h and v₂ = 24 km/h. The time difference Δ t is the sum of being 6 minutes early and 5 minutes late, which is 6 + 5 = 11 minutes. To keep the units consistent with km/h, the time difference must be converted to hours: Δ t = ¹¹⁄₆₀ hours.

Substituting these values into the formula: D = (30 × 24)/(30 - 24) × ¹¹⁄₆₀ D = ⁷²⁰⁄₆ × ¹¹⁄₆₀ D = 120 × ¹¹⁄₆₀ D = 2 × 11 = 22 km

Explanation

The compound interest formula is A = P(1 + r)ᵗ, where A is the final amount, P is the principal, r is the rate of interest, and t is the time in years. According to the first condition, the sum becomes eight times in 3 years, which is expressed as 8P = P(1 + r)³. Dividing both sides by P gives 8 = (1 + r)³. Since 8 = 2³, we can write 2³ = (1 + r)³, which simplifies to 1 + r = 2.

To find the time required for the sum to become sixteen times, we set up the equation 16P = P(1 + r)ᵗ. Substituting the previously determined value of 1 + r = 2 into this equation, we get 16 = 2ᵗ. Since 16 is 2⁴, the equation becomes 2⁴ = 2ᵗ. By comparing the exponents, we determine that t = 4 years.

Explanation

To solve the equation √(1 + ˣ⁄₉) = ¹³⁄₃, start by squaring both sides to remove the radical sign:

1 + ˣ⁄₉ = (¹³⁄₃)²

Squaring the fraction on the right-hand side:

1 + ˣ⁄₉ = ¹⁶⁹⁄₉

Next, subtract 1 from both sides to isolate the variable term. Representing 1 as ⁹⁄₉ allows for direct subtraction of the fractions:

ˣ⁄₉ = ¹⁶⁹⁄₉ - ⁹⁄₉ ˣ⁄₉ = ¹⁶⁰⁄₉

Finally, multiply both sides of the equation by 9 to solve for x:

x = ¹⁶⁰⁄₉ × 9 x = 160

Explanation

Using the algebraic identity (x + y)² = x² + y² + 2xy, the sum of the squares of two numbers x and y can be found using the formula x² + y² = (x + y)² - 2xy. Given that the sum x + y = 24 and the product xy = 143, substituting these values into the equation gives x² + y² = 24² - 2(143). Since 24² = 576 and 2 × 143 = 286, the calculation becomes 576 - 286 = 290.

Explanation

The expression 5⁷¹ + 5⁷² + 5⁷³ can be simplified by factoring out the lowest power of the base, which is 5⁷¹.

5⁷¹ + 5⁷² + 5⁷³ = 5⁷¹(5⁰ + 5¹ + 5²)

Simplifying the terms inside the parentheses: 5⁷¹(1 + 5 + 25) = 5⁷¹ × 31

To find the divisor among the given options, we can take one factor of 5 from 5⁷¹ and multiply it with 31: 5⁷⁰ × 5¹ × 31 = 5⁷⁰ × 155

Since 155 is a factor of the product 5⁷⁰ × 155, the total sum is completely divisible by 155.

Explanation

Let the two numbers be 10a and 10b, where a and b are co-prime integers because their H.C.F. is given as 10. The L.C.M. of 10a and 10b is 10ab. Given that the L.C.M. is 120, we have 10ab = 120, which simplifies to ab = 12.

To find the possible values for a and b, we look for pairs of factors of 12 that are co-prime. The possible co-prime pairs (a, b) are (1, 12) and (3, 4). If we take the pair (3, 4), the actual numbers are 10 × 3 = 30 and 10 × 4 = 40. The sum of these two numbers is 30 + 40 = 70. This matches one of the provided options.

Explanation

According to the property of remainders, if a number n leaves a remainder r when divided by a divisor d, then n² leaves the same remainder as r² when divided by d. In this case, n ≡ 2 5. Squaring both sides of the congruence gives n² ≡ 2² 5, which simplifies to n² ≡ 4 5. This can also be verified by expressing n as n = 5k + 2. Squaring this expression yields n² = (5k + 2)² = 25k² + 20k + 4. Since the terms 25k² and 20k are multiples of 5, the entire expression 5(5k² + 4k) + 4 leaves a remainder of 4 when divided by 5.

Explanation

Let x be the number. According to the student's calculation, adding 12 to the number and then dividing by 6 resulted in 112. This is represented by the equation (x + 12)/(6) = 112.

Multiplying both sides by 6, we find: x + 12 = 112 × 6 = 672 x = 672 - 12 = 660

The original instruction was to divide the number by 6 and add 12 to the quotient. Substituting x = 660 into this correct sequence gives: ˣ⁄₆ + 12 = ⁶⁶⁰⁄₆ + 12 110 + 12 = 122

Explanation

The runners will meet at the starting point for the first time at a time that is the Least Common Multiple (LCM) of their individual lap times. The given lap times are 200, 300, 360, and 450 seconds.

To find the LCM(200, 300, 360, 450), we use prime factorization: 200 = 2³ × 5² 300 = 2² × 3¹ × 5² 360 = 2³ × 3² × 5¹ 450 = 2¹ × 3² × 5²

The LCM is the product of the highest powers of all prime factors present across the numbers: LCM = 2ᵐᵃˣ⁽³^,²^,³^,¹⁾ × 3ᵐᵃˣ⁽⁰^,¹^,²^,²⁾ × 5ᵐᵃˣ⁽²^,²^,¹^,²⁾ LCM = 2³ × 3² × 5² LCM = 8 × 9 × 25 LCM = 1800