Quantitative Aptitude and Reasoning Archives

21. A father’s age is 2 years more than 4 times the age of his son. His ag

A father’s age is 2 years more than 4 times the age of his son. His age is also 2 years more than 5 times the age of his daughter. The average age of the father, the son and the daughter is 20 years. What is the age of the daughter ?

6 years

12 years

10 years

8 years

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Let F be the father’s age, S be the son’s age, and D be the daughter’s age.
According to the first statement: F = 4S + 2
According to the second statement: F = 5D + 2
According to the third statement: (F + S + D) / 3 = 20, which simplifies to F + S + D = 60.

From the first two equations, we can equate the expressions for F:
4S + 2 = 5D + 2
4S = 5D
S = 5D / 4

Now substitute the expressions for F and S in terms of D into the average age equation:
(5D + 2) + (5D / 4) + D = 60

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Multiply the entire equation by 4 to eliminate the fraction:
4(5D + 2) + 5D + 4D = 240
20D + 8 + 5D + 4D = 240
(20 + 5 + 4)D + 8 = 240
29D + 8 = 240
29D = 240 – 8
29D = 232
D = 232 / 29
D = 8

The age of the daughter is 8 years.

– Translate the word problem into a system of linear equations.
– Use substitution to solve the system of equations.

We can verify the ages:
Daughter’s age (D) = 8 years.
Father’s age (F) = 5D + 2 = 5(8) + 2 = 40 + 2 = 42 years.
Son’s age (S) = (F – 2) / 4 = (42 – 2) / 4 = 40 / 4 = 10 years.
Check the average: (F + S + D) / 3 = (42 + 10 + 8) / 3 = 60 / 3 = 20 years. This matches the given information.

22. What is the smallest number, which when multiplied by 9 gives the prod

What is the smallest number, which when multiplied by 9 gives the product having the digit 5 only in all places ?

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9528395

61728395

12345675

59382716

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We are looking for the smallest number, say ‘x’, such that when multiplied by 9, the product consists only of the digit 5. Let the product be P. So, 9 * x = P, where P is a number like 5, 55, 555, 5555, etc.
For P to be divisible by 9, the sum of its digits must be divisible by 9. If P consists only of the digit 5, the sum of its digits is 5 multiplied by the number of digits. For this sum to be divisible by 9, the number of digits (all 5s) must be a multiple of 9.
The smallest number of 5s that is a multiple of 9 is 9. So, the smallest possible product P is 555,555,555 (nine 5s).
Now, we find x by dividing P by 9:
x = 555,555,555 / 9
x = 61,728,395
Comparing this with the given options, option B matches this value.

– A number is divisible by 9 if the sum of its digits is divisible by 9.
– To find the smallest such number ‘x’, the product P must be the smallest number consisting only of 5s that is divisible by 9.
– The smallest number of 5s required for the sum of digits to be divisible by 9 is nine 5s (since 9 * 5 = 45, which is divisible by 9).

The calculation 555,555,555 ÷ 9 can be done using long division or by recognizing patterns. Dividing a repdigit of ‘n’ digits by 9 follows a specific pattern. For example, 111…1 (n times) divided by 9 gives 0.111…1. Here, we have nine 5s. Dividing 111,111,111 by 9 gives 12,345,679. So, 555,555,555 = 5 * 111,111,111. Therefore, 555,555,555 / 9 = 5 * (111,111,111 / 9) = 5 * 12,345,679 = 61,728,395.

23. The digit in the unit place of the number (347)¹⁹² x (143)²⁰⁵ is :

The digit in the unit place of the number (347)¹⁹² x (143)²⁰⁵ is :

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The correct answer is D.

To find the unit digit of a product, we need to find the unit digit of each number being multiplied and then find the unit digit of their product. The unit digit of a number raised to a power depends only on the unit digit of the base and the exponent. The unit digits follow cycles.
For (347)¹⁹²: The unit digit of the base is 7. The cycle of unit digits for powers of 7 is 7, 9, 3, 1 (length 4). Divide the exponent 192 by the cycle length 4: 192 ÷ 4 = 48 with a remainder of 0. When the remainder is 0 (or the power is a multiple of the cycle length), the unit digit is the last digit in the cycle, which is 1 (corresponding to 7⁴). So, the unit digit of (347)¹⁹² is 1.
For (143)²⁰⁵: The unit digit of the base is 3. The cycle of unit digits for powers of 3 is 3, 9, 7, 1 (length 4). Divide the exponent 205 by the cycle length 4: 205 ÷ 4 = 51 with a remainder of 1. The unit digit is the first digit in the cycle, which is 3 (corresponding to 3¹). So, the unit digit of (143)²⁰⁵ is 3.
The unit digit of (347)¹⁹² x (143)²⁰⁵ is the unit digit of the product of the individual unit digits: unit digit of (1 * 3) = unit digit of 3, which is 3.

The unit digit cycles for powers of digits 0-9 are:
0: 0 (length 1)
1: 1 (length 1)
2: 2, 4, 8, 6 (length 4)
3: 3, 9, 7, 1 (length 4)
4: 4, 6 (length 2)
5: 5 (length 1)
6: 6 (length 1)
7: 7, 9, 3, 1 (length 4)
8: 8, 4, 2, 6 (length 4)
9: 9, 1 (length 2)
For a power ‘n’, divide ‘n’ by the cycle length. If the remainder is ‘r’ (r > 0), the unit digit is the r-th digit in the cycle. If the remainder is 0, the unit digit is the last digit in the cycle.

24. The wheel of a car covers a distance of 44 km in 5000 revolutions. The

The wheel of a car covers a distance of 44 km in 5000 revolutions. The radius of the wheel is :

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14 cm

7 cm

28 cm

25 cm

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The correct answer is A.

The distance covered by a wheel in one revolution is equal to its circumference (C = 2πr). The total distance covered is the number of revolutions multiplied by the circumference.
Total distance = 44 km = 44,000 meters.
Number of revolutions = 5000.
Circumference C = Total distance / Number of revolutions = 44,000 m / 5000 = 44/5 m = 8.8 m.
Using the formula for circumference C = 2πr:
8.8 m = 2 * (22/7) * r
8.8 = (44/7) * r
r = (8.8 * 7) / 44 = (88/10 * 7) / 44 = (2 * 44 / 10 * 7) / 44 = (2/10) * 7 = 0.2 * 7 = 1.4 meters.
Converting meters to centimeters: 1.4 meters * 100 cm/meter = 140 cm.
However, 140 cm is not among the options. Let’s re-check the calculation assuming the distance was 4.4 km instead of 44 km, as the options are relatively small values in cm.
If distance = 4.4 km = 4400 m.
Circumference C = 4400 m / 5000 = 44/50 m = 22/25 m = 0.88 m.
Using C = 2πr:
0.88 m = 2 * (22/7) * r
0.88 = (44/7) * r
r = (0.88 * 7) / 44 = (88/100 * 7) / 44 = (2 * 7) / 100 = 14/100 = 0.14 meters.
Converting meters to centimeters: 0.14 meters * 100 cm/meter = 14 cm.
This matches option A. Given the options, it is highly probable that the intended distance in the question was 4.4 km, not 44 km.

Assuming a typo in the question is necessary to match the provided options. In competitive exams, if the exact calculation doesn’t yield any option, one might consider potential typos or approximations. In this case, a factor of 10 difference in distance leads to a factor of 10 difference in radius, making 14 cm the most likely intended answer corresponding to a distance of 4.4 km.

25. A climber tries to reach the top of an object which is 130 m high. He

A climber tries to reach the top of an object which is 130 m high. He can climb one metre per minute. However, as soon as he completes 5 metres, he slips down by one metre. The climber will reach the top in :

2 hours 10 minutes

2 hours 45 minutes

2 hours 42 minutes

2 hours 40 minutes

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The correct answer is C.

The climber climbs 1 meter per minute. He climbs 5 meters in 5 minutes and then slips 1 meter. So, in a sequence of climbing 5m followed by a slip, the net gain in height is 5m – 1m = 4m, and this takes 5 minutes of climbing time. The climber does not slip down if he reaches the top before completing a 5m segment *from his last slipped position*. The target height is 130 m.
The climber makes a net progress of 4m every 5 minutes of climbing until the point where the final climb to the top begins. Let’s find out how many 4m segments are needed to get close to 130m. The climber reaches heights that are multiples of 4m after each full cycle of climbing 5m and slipping 1m. The last segment climbed to reach the top will be less than or equal to 5m, and no slip will occur upon reaching 130m.
Consider the height reached after N full cycles (ending with a slip): H = 4N meters, taking 5N minutes. The final climb starts from H and goes up to 130m. This final climb must be <= 5m (from H). So, 130 - H <= 5, which means H >= 125. We need the smallest multiple of 4 that is greater than or equal to 125. 4 * 31 = 124 (not >= 125). 4 * 32 = 128. So, N = 32 cycles are completed.
After 32 cycles, the height reached is 128 m (after the 32nd slip). Time taken for 32 cycles = 32 * 5 minutes = 160 minutes.
From 128 m, the climber needs to climb 130 m – 128 m = 2 meters. This climb takes 2 minutes (at 1 m/min). Since 2m is less than 5m, no slip occurs upon reaching 130m.
Total time = Time for 32 cycles + Time for final climb = 160 minutes + 2 minutes = 162 minutes.
162 minutes = 2 hours and 42 minutes (120 minutes + 42 minutes).

This type of problem is a classic example of a linear progression with a setback, where special care must be taken for the final steps that complete the task. The key is to calculate the number of cycles required to reach a point from which the remaining climb is less than the distance that triggers a slip.

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26. In a party, each person takes at least one beverage. There are three b

In a party, each person takes at least one beverage. There are three beverages in the party – tea, coffee and milk. Each beverage is consumed by 30 persons. Five persons, who take tea, also take milk. Ten persons, who take milk, also take coffee. However, no person takes tea and coffee both. How many persons are there in the party ?

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The correct answer is 75 persons. This can be determined using the principle of inclusion-exclusion for sets, given the information about beverage consumption.

– Let T, C, and M represent the sets of people who take Tea, Coffee, and Milk, respectively.
– We are given: |T| = 30, |C| = 30, |M| = 30.
– |T ∩ M| = 5 (Persons who take tea and milk).
– |M ∩ C| = 10 (Persons who take milk and coffee).
– |T ∩ C| = 0 (No person takes tea and coffee both).
– Since no one takes tea and coffee together (|T ∩ C| = 0), it implies that no one takes tea, coffee, *and* milk together (|T ∩ C ∩ M| = 0).
– Each person takes at least one beverage, so the total number of persons in the party is the size of the union of the three sets: |T ∪ C ∪ M|.
– The Principle of Inclusion-Exclusion for three sets is:
|T ∪ C ∪ M| = |T| + |C| + |M| – (|T ∩ C| + |T ∩ M| + |C ∩ M|) + |T ∩ C ∩ M|
– Substitute the given values:
|T ∪ C ∪ M| = 30 + 30 + 30 – (0 + 5 + 10) + 0
|T ∪ C ∪ M| = 90 – 15 + 0
|T ∪ C ∪ M| = 75.

This problem can also be visualized using a Venn diagram. Starting with the intersections, we know |T ∩ C| = 0. Given |T ∩ M| = 5 and |M ∩ C| = 10, and |T ∩ C ∩ M| = 0, the people taking exactly two beverages are:
– Tea and Milk only: 5 – 0 = 5
– Milk and Coffee only: 10 – 0 = 10
– Tea and Coffee only: 0
People taking only one beverage:
– Tea only: |T| – (|T ∩ M| + |T ∩ C|) + |T ∩ C ∩ M| = 30 – (5 + 0) + 0 = 25
– Coffee only: |C| – (|C ∩ M| + |C ∩ T|) + |T ∩ C ∩ M| = 30 – (10 + 0) + 0 = 20
– Milk only: |M| – (|M ∩ T| + |M ∩ C|) + |T ∩ C ∩ M| = 30 – (5 + 10) + 0 = 15
Total persons = (Tea only) + (Coffee only) + (Milk only) + (Tea & Milk only) + (Milk & Coffee only) + (Tea & Coffee only) + (All three)
Total = 25 + 20 + 15 + 5 + 10 + 0 + 0 = 75.

27. A wire is in the shape of a circle. It is straightened and reshaped in

A wire is in the shape of a circle. It is straightened and reshaped into a rectangle whose sides are in the ratio 6 : 7. If the area of the rectangle is 4200 sq. cm, the radius of circle is, approximately :

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40 cm

42 cm

41 cm

35 cm

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The correct answer is approximately 41 cm. The radius of the circle is determined by equating its circumference to the perimeter of the rectangle, which is derived from the rectangle’s area and side ratio.

– The wire is reshaped, meaning the perimeter of the circle is equal to the perimeter of the rectangle.
– Rectangle sides are in the ratio 6:7. Let the sides be 6x and 7x.
– Area of rectangle = length * width = (6x)(7x) = 42x².
– Given area = 4200 sq. cm.
– 42x² = 4200 => x² = 4200 / 42 = 100 => x = 10 cm.
– Sides of the rectangle are 6 * 10 = 60 cm and 7 * 10 = 70 cm.
– Perimeter of rectangle = 2 * (length + width) = 2 * (60 + 70) = 2 * 130 = 260 cm.
– Circumference of the circle = 2πr, where r is the radius.
– Perimeter of circle = Perimeter of rectangle => 2πr = 260 cm.
– πr = 130 cm.
– r = 130 / π.
– Using the approximation π ≈ 3.14: r ≈ 130 / 3.14 ≈ 41.40 cm.
– Using the approximation π ≈ 22/7: r ≈ 130 / (22/7) = 130 * 7 / 22 = 910 / 22 = 455 / 11 ≈ 41.36 cm.
– The value approximately matches 41 cm.

This problem connects the concepts of area and perimeter for different geometric shapes. The key insight is that when a wire is reshaped, its total length (which corresponds to the perimeter of the original shape and the new shape) remains constant.

28. In a swimming competition, the players were asked to swim 9 km upstrea

In a swimming competition, the players were asked to swim 9 km upstream and then swim back to the point of starting. The winner of the competition could complete the task in 1 hour and 30 minutes. If the speed of the current of the river water was 8 km/hr, the speed of the winner is :

16 km/hr

20 km/hr

14 km/hr

18 km/hr

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The correct answer is 16 km/hr. This is the speed of the winner (swimmer) in still water, calculated based on the total time taken for the upstream and downstream journey.

– Let the speed of the swimmer in still water be ‘s’ km/hr.
– Speed of the current = 8 km/hr.
– Upstream speed (against the current) = (s – 8) km/hr.
– Downstream speed (with the current) = (s + 8) km/hr.
– Distance upstream = 9 km.
– Distance downstream = 9 km.
– Total time = 1 hour 30 minutes = 1.5 hours.
– Time upstream = Distance / Upstream speed = 9 / (s – 8) hours.
– Time downstream = Distance / Downstream speed = 9 / (s + 8) hours.
– Total time equation: 9/(s – 8) + 9/(s + 8) = 1.5
– Multiply the equation by (s – 8)(s + 8): 9(s + 8) + 9(s – 8) = 1.5(s – 8)(s + 8)
– 9s + 72 + 9s – 72 = 1.5(s² – 64)
– 18s = 1.5s² – 96
– Multiply by 2: 36s = 3s² – 192
– Divide by 3: 12s = s² – 64
– Rearrange into a quadratic equation: s² – 12s – 64 = 0
– Factor the equation: (s – 16)(s + 4) = 0
– Possible values for s are 16 or -4. Since speed must be positive, s = 16 km/hr.
– The speed of the swimmer (16 km/hr) must be greater than the speed of the current (8 km/hr) for upstream movement to be possible, which is true (16 > 8).

This is a standard ‘boats and streams’ problem. The crucial concepts are how the speed of the current affects the speed of the swimmer (or boat) upstream and downstream, and setting up the equation based on the total time taken for the round trip.

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29. Two trains are approaching each other on parallel tracks. Their length

Two trains are approaching each other on parallel tracks. Their lengths are 700 m and 400 m, respectively. The speed of the first train is 95 km/hr and that of the second train is 125 km/hr. The time required by the trains to cross each other is :

20 seconds

18 seconds

16 seconds

14 seconds

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The correct answer is 18 seconds. This is calculated by considering the relative speed of the trains approaching each other and the total distance they need to cover to cross each other.

– When two objects move towards each other, their relative speed is the sum of their individual speeds.
– Speed of first train = 95 km/hr.
– Speed of second train = 125 km/hr.
– Relative speed = 95 + 125 = 220 km/hr.
– To convert km/hr to m/s, multiply by 5/18: 220 km/hr = 220 * (5/18) m/s = 1100/18 m/s = 550/9 m/s.
– The total distance the trains must cover to cross each other is the sum of their lengths.
– Length of first train = 700 m.
– Length of second train = 400 m.
– Total distance = 700 + 400 = 1100 m.
– Time taken = Total distance / Relative speed.
– Time = 1100 m / (550/9 m/s) = 1100 * (9/550) seconds = (1100/550) * 9 seconds = 2 * 9 seconds = 18 seconds.

This is a classic problem involving relative speed. The key is to understand that the ‘distance’ covered when trains cross is the sum of their lengths, and when moving towards each other, their speeds add up. If they were moving in the same direction, the relative speed would be the difference between their speeds.

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30. If α and β are the roots of the equation x² – 7x + 11 = 0, then the va

If α and β are the roots of the equation x² – 7x + 11 = 0, then the value of α³ + β³ is equal to :

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112

224

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The correct answer is 112.

The given quadratic equation is x² – 7x + 11 = 0.
Let the roots be α and β.
According to Vieta’s formulas, for a quadratic equation ax² + bx + c = 0, the sum of the roots is α + β = -b/a and the product of the roots is αβ = c/a.
For the given equation x² – 7x + 11 = 0 (where a=1, b=-7, c=11):
Sum of roots: α + β = -(-7)/1 = 7.
Product of roots: αβ = 11/1 = 11.
We need to find the value of α³ + β³.
We can use the algebraic identity for the sum of cubes: α³ + β³ = (α + β)(α² – αβ + β²).
We can express α² + β² in terms of (α + β) and αβ:
α² + β² = (α + β)² – 2αβ.
Substituting this into the identity:
α³ + β³ = (α + β)[((α + β)² – 2αβ) – αβ]
α³ + β³ = (α + β)((α + β)² – 3αβ).
Now, substitute the values we found for (α + β) and αβ:
α + β = 7
αβ = 11
α³ + β³ = (7)((7)² – 3 × 11)
α³ + β³ = 7(49 – 33)
α³ + β³ = 7(16)
α³ + β³ = 112.

Vieta’s formulas provide a relationship between the coefficients of a polynomial and the sums and products of its roots. For a quadratic equation ax² + bx + c = 0, the formulas are: sum of roots = -b/a, product of roots = c/a. These are very useful for solving problems involving symmetric expressions of roots without explicitly finding the roots themselves.