RRB NTPC2021MathematicsPage 2 of 36

RRB NTPC CBT 1 2021 Mathematics Questions with Solutions

Questions from every shift of the year, grouped by section. Questions 51–100 of 1,785, each with the correct answer and a step-by-step explanation.

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51Mathematics
If cos x + (1)/(cos x) = 2 then find the value of cosⁿx + (1)/(cosⁿx)
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Answer: (D) 2

Explanation

Given the equation cos x + 1/cos x = 2, let cos x = t.

t + 1/t = 2 t² - 2t + 1 = 0 (t - 1)² = 0 t = 1

Thus, cos x = 1.

Substituting cos x = 1 into the required expression: cosⁿ x + 1/cosⁿ x = 1ⁿ + 1/1ⁿ = 1 + 1 = 2

52Mathematics
If one angle of a parallelogram is 48° less than twice the smallest angle, then the measure of the largest angle of the parallelogram will be:
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Answer: (A) 104°

Explanation

Adjacent angles of a parallelogram are supplementary, summing to 180°.

Let the smallest angle be x. The adjacent angle is given as 2x - 48°.

x + (2x - 48°) = 180° 3x = 228° x = 76°

The largest angle = 180° - 76° = 104°

53Mathematics
A person invested ⅔ of his capital at the rate of 6% ⅕ at the rate of 10% and the remainder at the rate of 15% . If his annual income is ₹600, the capital will be:
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Answer: (D) ₹7,500

Explanation

Let total capital be C.

The remaining fraction of capital = 1 - (2/3 + 1/5) = 1 - 13/15 = 2/15.

The effective overall rate of interest is: (2/3 × 6%) + (1/5 × 10%) + (2/15 × 15%) = 4% + 2% + 2% = 8%

Since total annual income is 600: 8% of C = 600 C = (600 × 100) / 8 = 7,500

54Mathematics
Find the sum of the numbers between 400 and 600 such that when they are divided by 6, 12 and 16, there will be no remainder.
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Answer: (D) 2016

Explanation

Numbers divisible by 6, 12, and 16 must be multiples of their Least Common Multiple (LCM). LCM(6, 12, 16) = 48

The multiples of 48 between 400 and 600 are: 48 × 9 = 432 48 × 10 = 480 48 × 11 = 528 48 × 12 = 576

Using the formula for the sum of an arithmetic progression: Sum = n/2 × (a + l) where n = 4 terms, first term a = 432, and last term l = 576.

Sum = 4/2 × (432 + 576) = 2 × 1008 = 2016.

55Mathematics
The LCM of √(36) and √(64) is:
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Answer: (D) 24

Explanation

Simplifying the given terms: √36 = 6 √64 = 8

To find the LCM of 6 and 8, write their prime factorizations: 6 = 2 × 3 8 = 2³

LCM(6, 8) = 2³ × 3 = 8 × 3 = 24

56Mathematics
If a selling price of ₹36 results in a 10% discount on a list price, what selling price will result in a 20% discount on the list price?
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Answer: (C) 32

Explanation

Selling Price = List Price × (100 - Discount%)/100

Since the list price remains constant, the ratio of selling prices for two different discount rates is given by: SP₂ = SP₁ × (100 - D₂) / (100 - D₁)

Substituting SP₁ = ₹36, D₁ = 10%, and D₂ = 20%: SP₂ = 36 × (100 - 20) / (100 - 10) SP₂ = 36 × 80 / 90 SP₂ = ₹32

57Mathematics
In a class of 100 students, the mean marks obtained in a certain subject is 25 and in another class of 50 students, the mean marks obtained in the same subject is 70. The mean marks obtained by the students of both the classes taken together is:
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Answer: (C) 40

Explanation

The combined mean x̄ of two groups with sample sizes n₁, n₂ and means x̄₁, x̄₂ is given by the weighted mean formula:

x̄ = (n₁x̄₁ + n₂x̄₂) / (n₁ + n₂)

Given: n₁ = 100, x̄₁ = 25 n₂ = 50, x̄₂ = 70

Substituting the values: x̄ = (100 × 25 + 50 × 70) / (100 + 50) x̄ = (2500 + 3500) / 150 x̄ = 6000 / 150 = 40

58Mathematics
If a, b, c, d are four numbers, such that the LCM of a and b is b, the LCM of b and c is c and the LCM of c and d is d, then the LCM of a, b, c and d will be:
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Answer: (A) d

Explanation

If LCM(x, y) = y, then x is a factor of y.

Given that LCM(a, b) = b, a divides b. Since LCM(b, c) = c, b divides c. Since LCM(c, d) = d, c divides d.

By transitivity of divisibility, a, b, and c are all factors of d. Since d is divisible by a, b, c, and itself, the Least Common Multiple of all four numbers a, b, c, and d is d.

59Mathematics
The price of mobile is first decreased by 20% and then increased by 10% . The net change in the price will be:
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Answer: (D) 12%

Explanation

Using the successive percentage change formula:

Net % change = a + b + (a × b) / 100

Substituting a = -20% (decrease) and b = +10% (increase):

Net % change = -20 + 10 + (-20 × 10) / 100 = -10 - 2 = -12%

The negative sign indicates an overall decrease, giving a net change of 12%.

60Mathematics
A man travelled from a school to a park at a speed of 20km / h and walked back at a speed of 5km / h If the whole journey took 5h find the distance of the school from the park.
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Answer: (D) 20km

Explanation

When a distance d is traveled at speed v₁ and returned at speed v₂, the total time T taken for the entire journey is given by:

T = d/v₁ + d/v₂

Substituting the given values v₁ = 20 km/h, v₂ = 5 km/h, and T = 5 h:

5 = d/20 + d/5

5 = (d + 4d) / 20

5 = 5d / 20

5 = d / 4

d = 20 km

61Mathematics
If a tan α = b, then the value of cos α will be:
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Answer: (C) (a)/(√(a² + b²))

Explanation

Given a tan α = b, we have tan α = b/a.

Using the right-triangle definition of trigonometric functions, if tan α = Perpendicular / Base = b/a, then Base = a and Perpendicular = b.

By the Pythagorean theorem: Hypotenuse = √(Base² + Perpendicular²) = √(a² + b²)

Therefore, the cosine of α is given by: cos α = Base / Hypotenuse = a / √(a² + b²)

62Mathematics
The given table shows the details of the expenditure incurred (in ₹ crores) by a tea garden during the years 2013 to 2017. Study the table carefully and answer the following question. Year | Salary | Fuel etc | Bonus | Interest on loans | Taxes 2013 | 60.5 | 20.0 | 0.8 | 4.88 | 18.6 2014 | 68.4 | 23.4 | 0.4 | 7.00 | 22.6 2015 | 62.8 | 19.4 | 0.86 | 8.85 | 16.8 2016 | 67.2 | 28.8 | 0.89 | 6.28 | 17.5 2017 | 84.0 | 30.2 | 0.96 | 10.11 | 21.6 In which year was the total expenditure on 'Taxes' and 'Interest on loans' less than the expenditure on 'Fuel etc'?
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Answer: (C) 2016

Explanation

In 2016, the expenditure on 'Fuel etc' was ₹28.8 crores, while the total expenditure on 'Taxes' and 'Interest on loans' was ₹17.5 + ₹6.28 = ₹23.78 crores. Here, ₹23.78 crores is less than ₹28.8 crores.

63Mathematics
The given table shows the details of the expenditure incurred (in ₹ crores) by a tea garden from 2013 to 2017. Study the table carefully and answer the following question. Year | Salary | Fuel etc | Bonus | Interest on loans | Taxes 2013 | 60.5 | 20.0 | 0.8 | 4.88 | 18.6 2014 | 68.4 | 23.4 | 0.4 | 7.00 | 22.6 2015 | 62.8 | 19.4 | 0.86 | 8.85 | 16.8 2016 | 67.2 | 28.8 | 0.89 | 6.28 | 17.5 2017 | 84.0 | 30.2 | 0.96 | 10.11 | 21.6 The average yearly expenditure on 'Salary' from 2013 to 2017 was approximately:
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Answer: (B) ₹68.58 crore

Explanation

Average expenditure = Sum of expenditure on Salary from 2013 to 2017 / Total number of years

Sum of Salary expenditure = 60.5 + 68.4 + 62.8 + 67.2 + 84.0 = 342.9 crore

Average = 342.9 / 5 = ₹68.58 crore

64Mathematics
The given table shows the details of the expenditure incurred (in ₹ crores) by a tea garden from 2013 to 2017. Study the table carefully and answer the following question. Year | Salary | Fuel etc | Bonus | Interest on loans | Taxes 2013 | 60.5 | 20.0 | 0.8 | 4.88 | 18.6 2014 | 68.4 | 23.4 | 0.4 | 7.00 | 22.6 2015 | 62.8 | 19.4 | 0.86 | 8.85 | 16.8 2016 | 67.2 | 28.8 | 0.89 | 6.28 | 17.5 2017 | 84.0 | 30.2 | 0.96 | 10.11 | 21.6 The range of expenditure incurred over the five-year period is most similar for which of the following items of expenditure?
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Answer: (D) Interest on Loans and Taxes

Explanation

Range is defined as the difference between the maximum and minimum expenditure values over the given period.

Range for Interest on loans = 10.11 - 4.88 = 5.23 ₹ crores Range for Taxes = 22.6 - 16.8 = 5.8 ₹ crores

The calculated ranges of 5.23 ₹ crores for Interest on loans and 5.8 ₹ crores for Taxes are closest to each other, representing the most similar range of expenditure.

65Mathematics
The given table shows the details of the expenditure incurred (in C crores) by a tea garden from 2013 to 2017. Study the table carefully and answer the following question. Year Items of Expenditure (in Crores) Salary Fuel etc Bonus Interest on loans 2013 60.5 20.0 0.8 4.88 2014 68.4 23.4 0.4 7.00 2015 62.8 19.4 0.86 8.85 2016 67.2 28.8 0.89 6.28 2017 84.0 30.2 0.96 10.11 In which year was the percentage increase of the expenditure on salary, as compared to the previous year, more than 20% ?
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Answer: (D) 2017

Explanation

Percentage Increase = [(Salary in current year - Salary in previous year) / Salary in previous year] × 100

For 2017: Percentage increase = [(84.0 - 67.2) / 67.2] × 100 = (16.8 / 67.2) × 100 = 25%

Since 25% is greater than 20%, this satisfies the given condition.

66Mathematics
How will you write 2.84 hours in hours, minutes and seconds?
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Answer: (D) 2 hours 50 minutes 24 seconds

Explanation

To convert 2.84 hours into hours, minutes, and seconds:

Separating the whole number part gives 2 hours.

The decimal part is converted to minutes by multiplying by 60: 0.84 × 60 = 50.4 minutes

Taking the whole number part gives 50 minutes.

The remaining decimal portion of minutes is converted to seconds by multiplying by 60: 0.4 × 60 = 24 seconds

Combining these yields 2 hours 50 minutes 24 seconds.

67Mathematics
If a² + b² + c² + d² = 1, what will be the maximum value of the product abcd?
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Answer: (A) ¹⁄₁₆

Explanation

By the Arithmetic Mean - Geometric Mean (AM-GM) inequality for the non-negative terms a², b², c², and d²:

(a² + b² + c² + d²) / 4 ≥ (a² × b² × c² × d²)⁽1/4)

Substituting a² + b² + c² + d² = 1:

1 / 4 ≥ (a²b²c²d²)⁽1/4)

Raising both sides to the fourth power:

(1 / 4)⁴ ≥ a²b²c²d² 1 / 256 ≥ (abcd)²

Taking the square root gives abcd ≤ 1/16, with equality holding when a = b = c = d = ½.

68Mathematics
Two quantities are in the ratio 3:5. If each quantity is decreased by 9 then the ratio becomes 5:9. Find the smaller quantity.
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Answer: (C) 54

Explanation

Let the two quantities be 3x and 5x, where 3x is the smaller quantity.

According to the given condition: (3x - 9) / (5x - 9) = 5 / 9

Cross-multiplying to solve for x: 9(3x - 9) = 5(5x - 9) 27x - 81 = 25x - 45 2x = 36 x = 18

The smaller quantity is 3x = 3 × 18 = 54.

69Mathematics
A daily wage labourer was engaged for a certain number of days for 5,850, but being absent on some of those days he was paid only 5,200. What was his maximum possible daily wage?
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Answer: (B) 650

Explanation

To find the maximum possible daily wage, we calculate the Highest Common Factor (HCF) of the total amount offered (₹5,850) and the actual amount received (₹5,200), as the daily wage must evenly divide both amounts.

HCF(5850, 5200) = HCF(5850 - 5200, 5200) = HCF(650, 5200)

Since 5,200 ÷ 650 = 8 and 5,850 ÷ 650 = 9, the greatest common divisor of both amounts is 650.

70Mathematics
If the largest 4- digit number is subtracted from the smallest 6- digit number, then the remainder will be:
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Answer: (A) 90001

Explanation

The smallest 6-digit number is 100000 and the largest 4-digit number is 9999.

Subtracting 9999 from 100000: 100000 - 9999 = 90001

71Mathematics
Richa purchased a car for ₹6,50,000 and sold it for ₹7,80,000. What is the percentage profit she made on the car?
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Answer: (A) 20%

Explanation

Profit is calculated by subtracting the Cost Price from the Selling Price: Profit = Selling Price - Cost Price Profit = ₹7,80,000 - ₹6,50,000 = ₹1,30,000

To find the percentage profit on the car, divide the total profit by the original cost price and multiply by 100: Percentage Profit = (Profit / Cost Price) × 100 Percentage Profit = (1,30,000 / 6,50,000) × 100 Percentage Profit = 1/5 × 100 = 20%

72Mathematics
If 6(sec²59° - cot²31°) + ⅔sin 90° - 3tan²56°ytan²34° = ʸ⁄₂ , then the value of y is:
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Answer: (B) 2

Explanation

Using trigonometric complementary angle identities and standard relations: sec 59° = cosec(90° - 59°) = cosec 31°, so sec²59° - cot²31° = cosec²31° - cot²31° = 1 sin 90° = 1 tan 56° = cot(90° - 56°) = cot 34°, so tan²56° × tan²34° = cot²34° × tan²34° = 1

Substituting these values into the given equation: 6(1) + (2/3)(1) - 3y(1) = y/3 6 + 2/3 - 3y = y/3 20/3 = 3y + y/3 20/3 = 10y/3 10y = 20 y = 2

73Mathematics
Find the compound interest on ₹1,200 at 12% p.a. for 6 months compounded quarterly.
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Answer: (C) ₹73.08

Explanation

For quarterly compounding, the rate of interest per quarter is r = 12% / 4 = 3%, and the time duration of 6 months corresponds to n = 6 / 3 = 2 quarters.

Using the compound interest formula: Amount (A) = P × (1 + r/100)ⁿ A = 1200 × (1 + 3/100)² = 1200 × (1.03)² = 1200 × 1.0609 = ₹1,273.08

Compound Interest (CI) = Amount - Principal CI = ₹1,273.08 - ₹1,200 = ₹73.08

74Mathematics
Rakesh donates blood twice in 3 years each time 330 ml. How many litres of blood will he donate in 6 years?
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Answer: (A) 1.32L

Explanation

Number of 3-year periods in 6 years = 6 ÷ 3 = 2

Total donations in 6 years = 2 × 2 = 4 times

Total blood donated = 4 × 330 ml = 1320 ml

Converting ml to litres: 1320 ml = 1320 ÷ 1000 = 1.32 L

75Mathematics
Find the area of an equilateral triangle with side 12 cm.
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Answer: (D) 36√3cm²

Explanation

The area of an equilateral triangle with side length a is given by the formula:

Area = (√3 / 4) × a²

Substituting side a = 12 cm into the formula:

Area = (√3 / 4) × 12² = (√3 / 4) × 144 = 36√3 cm²

76Mathematics
If the numbers from 1 to 26 that are divisible by 2 are arranged in descending order, which number will be at the 9th place from the bottom?
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Answer: (A) 18

Explanation

The numbers from 1 to 26 that are divisible by 2 are the even numbers starting from 2 up to 26.

When these numbers are arranged in descending order (26, 24, 22, ..., 2), the smallest number is at the bottom. Therefore, counting from the bottom is equivalent to taking the sequence of even numbers in ascending order starting from 2.

The nth number from the bottom is given by the formula: nth term = 2 × n

For the 9th place from the bottom (n = 9): 9th term = 2 × 9 = 18

77Mathematics
The area of a square is 289 cm² . Find the length of its diagonal.
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Answer: (A) 17√2 cm

Explanation

The area of a square is related to its side length by Area = a². a = √289 = 17 cm

The length of the diagonal of a square is given by d = a√2. Substituting the side length: d = 17√2 cm

78Mathematics
Find the least number of five digits, which is exactly divisible by 472.
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Answer: (B) 10384

Explanation

The smallest five-digit number is 10000. Dividing 10000 by 472 yields a quotient of 21 and a remainder of 88 ( since 10000 = 472 × 21 + 88 ). To find the smallest five-digit multiple of 472, we take the next integer quotient 22 and compute 472 × 22 = 10384. Alternatively, adding the difference between the divisor and the remainder (472 - 88 = 384) to 10000 gives 10384.

79Mathematics
Flooring of a room 12 m long and 8 m wide is to be designed by squares of maximum possible area. Find the number of square designs required.
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Answer: (B) 6

Explanation

To find the maximum possible area of the square designs, the side length of each square must be equal to the Highest Common Factor (HCF) of the dimensions of the room.

HCF(12, 8) = 4 m

Number of square designs required = (Length ÷ Side) × (Width ÷ Side) = (12 ÷ 4) × (8 ÷ 4) = 3 × 2 = 6

80Mathematics
Find the ratio of the measure of an angle of a regular pentagon to that of a regular octagon.
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Answer: (B) 4:5

Explanation

The measure of an interior angle of a regular polygon with n sides is given by the formula ((n - 2) × 180°) / n.

For a regular pentagon (n = 5): Interior angle = ((5 - 2) × 180°) / 5 = (3 × 180°) / 5 = 108°

For a regular octagon (n = 8): Interior angle = ((8 - 2) × 180°) / 8 = (6 × 180°) / 8 = 135°

Ratio of the interior angle of a pentagon to an octagon = 108 : 135 = 4 : 5.

81Mathematics
The diameter of a wheel is 88 cm. Find the number of revolutions in which it will cover a distance of 8712 m. (Use π = ²²⁄₇ )
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Answer: (C) 3150

Explanation

Circumference of the wheel = π × diameter = (22/7) × 88 cm = 1936/7 cm = 1936/700 m

Distance covered in one revolution equals the circumference of the wheel.

Number of revolutions = Total distance ÷ Circumference = 8712 ÷ (1936 / 700) = (8712 × 700) ÷ 1936 = 4.5 × 700 = 3150

82Mathematics
A man's working hours per day were increased by 35% and his wages per hour were decreased by 25%. By what percentage were his daily earnings increased?
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Answer: (C) 1.25

Explanation

Daily earnings equal the product of working hours per day and wages per hour.

Using the net percentage change formula for the product of two quantities: Net % change = a + b + (a × b) / 100

Given a = +35% and b = -25%: Net % change = 35 - 25 + (35 × (-25)) / 100 = 10 - 875 / 100 = 10 - 8.75 = 1.25%

83Mathematics
Simplify. (6.25 + ⁵⁄₇ × 28 - 5)/(¾ × (15.8 - 3.4) + 5 × 2.39)
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Answer: (B) 1

Explanation

Simplify the numerator and denominator separately:

Numerator = 6.25 + (5/7) × 28 − 5 = 6.25 + 20 − 5 = 21.25

Denominator = (3/4) × (15.8 − 3.4) + 5 × 2.39 = (3/4) × 12.4 + 11.95 = 9.3 + 11.95 = 21.25

Dividing the numerator by the denominator: 21.25 / 21.25 = 1

84Mathematics
A scooterist is travelling at the speed of 50km / h. How much distance will he cover in 2½ h?
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Answer: (B) 125km

Explanation

Distance = Speed × Time

Given: Speed = 50 km/h Time = 2½ h = 2.5 h

Distance = 50 × 2.5 = 125 km

85Mathematics
A dog saw a cat at a distance of 280m. The cat at once ran with the speed of 10km / h and the dog also ran to catch it with the speed of 24km / h. How much time will the dog take to catch the cat?
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Answer: (B) 1.2min

Explanation

Relative Speed = Speed of dog - Speed of cat Relative Speed = 24 km/h - 10 km/h = 14 km/h

Converting relative speed to meters per minute: Relative Speed = (14 × 1000) / 60 = 700/3 m/min

Time = Distance / Relative Speed Time = 280 / (700/3) = (280 × 3) / 700 = 1.2 minutes

86Mathematics
What is the sum of the squares of the numbers from 1 to 12?
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Answer: (D) 650

Explanation

The sum of the squares of the first n natural numbers is given by the formula S = n(n + 1)(2n + 1) / 6.

For n = 12: S = 12 × (12 + 1) × (2 × 12 + 1) / 6 S = 12 × 13 × 25 / 6 S = 2 × 13 × 25 = 650

87Mathematics
The LCM of ⅓, ⁷⁄₆, ⁵⁄₉, ⁴⁄₂₇, ⁸⁄₁₅ is:
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Answer: (B) ²⁸⁰⁄₃

Explanation

The LCM of a set of fractions is given by the formula: LCM of fractions = LCM of numerators / HCF of denominators

For the given fractions 1/3, 7/6, 5/9, 4/27, and 8/15:

Numerators = 1, 7, 5, 4, 8 LCM(1, 7, 5, 4, 8) = 280

Denominators = 3, 6, 9, 27, 15 HCF(3, 6, 9, 27, 15) = 3

Substituting these values into the formula yields 280 / 3.

88Mathematics
The HCF of two numbers is 19 and the other two factors of their LCM are 11 and 13. The larger of the two numbers is:
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Answer: (C) 247

Explanation

Let the two numbers be 19a and 19b, where a and b are co-prime factors of the LCM. The problem states that the other two factors of their LCM are 11 and 13, so a = 11 and b = 13.

The larger of the two numbers is obtained by multiplying the HCF by the larger co-prime factor:

Larger number = 19 × 13 = 247

89Mathematics
Find the value of the following. cos 15° - sin 75° cos 15° + sin 75°
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Answer: (B) 0

Explanation

Using the complementary angle identity sin(90° - θ) = cos θ, sin 75° can be written as sin(90° - 15°) = cos 15°.

Substituting sin 75° = cos 15° into the given expression (cos 15° - sin 75°) / (cos 15° + sin 75°) gives:

Numerator = cos 15° - cos 15° = 0

Denominator = cos 15° + cos 15° = 2 cos 15°

Since the numerator evaluates to 0 and the denominator is non-zero, the value of the entire expression is 0 / (2 cos 15°) = 0.

90Mathematics
If the sum of two numbers is 84 and their HCF and LCM are 3 and 124 respectively, the sum of the reciprocals of the two numbers will be:
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Answer: (D) 7/31

Explanation

Let the two numbers be x and y.

The sum of two numbers is given as x + y = 84.

The product of two numbers is equal to the product of their HCF and LCM: x × y = HCF × LCM = 3 × 124 = 372

The sum of the reciprocals of the two numbers is: 1/x + 1/y = (x + y) / (x × y)

Substituting the values: 1/x + 1/y = 84 / 372 = 7 / 31

91Mathematics
The area of an isosceles right angle triangle is 81 cm². Find the length of its hypotenuse.
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Answer: (B) 18 cm

Explanation

The area of an isosceles right-angled triangle with hypotenuse h is given by the formula Area = h² / 4.

Substituting the given area of 81 cm² into the formula: 81 = h² / 4 h² = 324 h = √324 = 18 cm

92Mathematics
In a polygon, the sum of the interior angles is triple the sum of the exterior angles. The number of sides is:
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Answer: (A) 8

Explanation

For an n-sided polygon, the sum of interior angles is given by (n - 2) × 180° and the sum of exterior angles is always 360°.

According to the problem: (n - 2) × 180° = 3 × 360° (n - 2) × 180° = 1080° n - 2 = 6 n = 8

93Mathematics
35% of a number is the same as 30% of another number. Find the ratio of the first number to the second number.
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Answer: (B) 6 : 7

Explanation

Let the first number be x and the second number be y.

35% of x = 30% of y 35x = 30y x/y = 30/35 = 6/7

Thus, x : y = 6 : 7.

94Mathematics
Angle 54° is equivalent to (in radians):
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Answer: (C) 3π/10

Explanation

To convert an angle from degrees to radians, multiply the degree measure by π / 180°.

Radians = 54° × (π / 180°) = 54π / 180

Dividing both the numerator and the denominator by 18 yields 3π/10.

95Mathematics
The sum of two numbers is 80 and their difference is 8. The ratio of the first number to the second number will be:
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Answer: (A) 11:9

Explanation

Let the two numbers be x and y. Given, x + y = 80 and x - y = 8.

Adding the two equations gives: 2x = 88 ⇒ x = 44

Subtracting the two equations gives: 2y = 72 ⇒ y = 36

The required ratio of the first number to the second number is: x : y = 44 : 36 = 11 : 9

96Mathematics
The selling price of 40 items is equal to the purchase price of 35 items. What is the percentage profit or loss?
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Answer: (A) 12½% loss

Explanation

Let CP be the cost price of one item and SP be the selling price of one item.

Given, 40 × SP = 35 × CP SP / CP = 35 / 40 = 7 / 8

Since SP is less than CP, it is a case of loss.

Loss percentage = [(CP - SP) / CP] × 100 Loss percentage = [(8 - 7) / 8] × 100 = (1 / 8) × 100 = 12½%

97Mathematics
A man completes a journey in 11 h. He travels first half of the journey at the speed of 25 km/h and the second half at the speed of 30 km/h. Find the total distance of the journey.
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Answer: (A) 300 km

Explanation

Average speed for two equal halves of a journey is given by the formula 2 × v₁ × v₂ / (v₁ + v₂).

Substituting v₁ = 25 km/h and v₂ = 30 km/h: Average speed = (2 × 25 × 30) / (25 + 30) = 1500 / 55 = 300 / 11 km/h

Total distance = Average speed × Total time Total distance = (300 / 11) × 11 = 300 km

98Mathematics
0.53̄2̄ is equivalent to the fraction:
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Answer: (D) 527/990

Explanation

To convert a mixed recurring decimal 0.53̄2̄ (where 32 repeats indefinitely) into a fraction, use the standard conversion formula:

Fraction = (Complete number after decimal - Non-recurring part) / (Number of 9s equal to repeating digits followed by 0s equal to non-repeating digits)

For 0.53̄2̄: Numerator = 532 - 5 = 527 Denominator = 990 (two 9s for '32' and one 0 for '5')

Algebraically: Let x = 0.5323232... 10x = 5.323232... 1000x = 532.323232... 1000x - 10x = 532 - 5 990x = 527 x = 527/990

99Mathematics
Study the given table and answer the question that follows. The table gives marks scored by 5 students in 5 subjects in an exam. Subject Jeeva Justin Hetal Kuber Alia English 76 88 54 72 90 Maths 82 85 78 64 60 Science 69 80 82 80 75 History 70 72 76 83 80 Geography 86 90 85 80 73 Find the average marks scored in science by the students.
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Answer: (D) 77.2

Explanation

Average marks in Science = Sum of marks scored by all students in Science / Number of students

Sum of Science marks = 69 + 80 + 82 + 80 + 75 = 386 Number of students = 5

Average = 386 / 5 = 77.2

100Mathematics
Study the given table and answer the question that follows. The table gives marks scored by 5 students in 5 subjects in an exam. Subject Jeeva Justin Hetal Kuber English 76 88 54 72 Maths 82 85 78 64 Science 69 80 82 80 History 70 72 76 83 Geography 86 90 85 80 Which student has scored highest marks in at least 2 subjects?
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Answer: (C) Justin

Explanation

Justin scored the highest marks in English (88), Maths (85), and Geography (90). Having secured the top score in three subjects, Justin is the student who scored the highest marks in at least 2 subjects.

Answer key

Tap a number to jump to that question.

  1. 51D
  2. 52A
  3. 53D
  4. 54D
  5. 55D
  6. 56C
  7. 57C
  8. 58A
  9. 59D
  10. 60D
  11. 61C
  12. 62C
  13. 63B
  14. 64D
  15. 65D
  16. 66D
  17. 67A
  18. 68C
  19. 69B
  20. 70A
  21. 71A
  22. 72B
  23. 73C
  24. 74A
  25. 75D
  26. 76A
  27. 77A
  28. 78B
  29. 79B
  30. 80B
  31. 81C
  32. 82C
  33. 83B
  34. 84B
  35. 85B
  36. 86D
  37. 87B
  38. 88C
  39. 89B
  40. 90D
  41. 91B
  42. 92A
  43. 93B
  44. 94C
  45. 95A
  46. 96A
  47. 97A
  48. 98D
  49. 99D
  50. 100C

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