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IPMAT Mock Test - 8 (New Pattern) - Commerce MCQ


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30 Questions MCQ Test IPMAT Mock Test Series - IPMAT Mock Test - 8 (New Pattern)

IPMAT Mock Test - 8 (New Pattern) for Commerce 2024 is part of IPMAT Mock Test Series preparation. The IPMAT Mock Test - 8 (New Pattern) questions and answers have been prepared according to the Commerce exam syllabus.The IPMAT Mock Test - 8 (New Pattern) MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for IPMAT Mock Test - 8 (New Pattern) below.
Solutions of IPMAT Mock Test - 8 (New Pattern) questions in English are available as part of our IPMAT Mock Test Series for Commerce & IPMAT Mock Test - 8 (New Pattern) solutions in Hindi for IPMAT Mock Test Series course. Download more important topics, notes, lectures and mock test series for Commerce Exam by signing up for free. Attempt IPMAT Mock Test - 8 (New Pattern) | 100 questions in 90 minutes | Mock test for Commerce preparation | Free important questions MCQ to study IPMAT Mock Test Series for Commerce Exam | Download free PDF with solutions
IPMAT Mock Test - 8 (New Pattern) - Question 1
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Answer the questions on the basis of the information given below:

A, B, C. D and E are five brothers. They have assets in the form of flats and cash only. Together they have 9 flats and each brother has at least 1 flat. Value (in terms of rupees) of all flats is equal and constant. Cash with each brother is an integral multiple of Rs. 10 lakhs. The amount of cash with B is just enough to buy one more flat, all the other brothers have cashless than that of B. E has got the maximum number of flats. A number of flats with A is equal to the sum of the number of flats with B and C. Amount of cash with B is double of that with D, but D has got assets worth Rs. 20 lakhs more than that of B.

What is the value of 1 flat?

Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 1
E has got the maximum number of flats and number of flats with A is equal to the sum of a number of flats with B and C. As there are 9 flats, so possible solutions are when E, A, B, C and D have 3, 2, 1, 1 and 2 or 4, 2, 1. 1 and 1 flats respectively.

Case I:B and D have one flat and two flats respectively.

Let the value of a flat be Rs. x. So, the amount of cash with B = x.

So total assets of B = 2x

Total assets of As D has got assets worth 20 lakhs more than that of B

Case II:

B and D have one flat each.

Let value of a flat be Rs. x. So, amount of cash with B = x

⇒ Total assets of B = 2x

Total assets of

As D has got assets worth 20 lakhs more than B

Here the value of x is coming as negative. So this case is not possible.

We can collate the final distribution of assets in the following table:

In above table B's flat is 40 lacks so,value of 1 flat is Rs 40 lacks.

IPMAT Mock Test - 8 (New Pattern) - Question 2
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Answer the questions on the basis of the information given below:

A, B, C. D and E are five brothers. They have assets in the form of flats and cash only. Together they have 9 flats and each brother has at least 1 flat. Value (in terms of rupees) of all flats is equal and constant. Cash with each brother is an integral multiple of Rs. 10 lakhs. The amount of cash with B is just enough to buy one more flat, all the other brothers have cashless than that of B. E has got the maximum number of flats. A number of flats with A is equal to the sum of the number of flats with B and C. Amount of cash with B is double of that with D, but D has got assets worth Rs. 20 lakhs more than that of B.

What is the ratio of the maximum possible value of assets of E to the minimum possible value of assets of C?

Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 2
E has got the maximum number of flats and number of flats with A is equal to the sum of a number of flats with B and C. As there are 9 flats, so possible solutions are when E, A, B, C and D have 3, 2, 1, 1 and 2 or 4, 2, 1. 1 and 1 flats respectively.

Case I:B and D have one flat and two flats respectively.

Let the value of a flat be Rs. x. So, the amount of cash with B = x.

So total assets of B = 2x

Total assets of As D has got assets worth 20 lakhs more than that of B

Case II:

B and D have one flat each.

Let value of a flat be Rs. x. So, amount of cash with B = x

⇒ Total assets of B = 2x

Total assets of

As D has got assets worth 20 lakhs more than B

Here the value of x is coming as negative. So this case is not possible.

We can collate the final distribution of assets in the following table:

Maximum possible assets of E = 150 lakhs and minimum possible assets of C = 50 lakhs. so, Ratio is = 3 : 1​.

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IPMAT Mock Test - 8 (New Pattern) - Question 3
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Answer the questions on the basis of the information given below:

A, B, C. D and E are five brothers. They have assets in the form of flats and cash only. Together they have 9 flats and each brother has at least 1 flat. Value (in terms of rupees) of all flats is equal and constant. Cash with each brother is an integral multiple of Rs. 10 lakhs. The amount of cash with B is just enough to buy one more flat, all the other brothers have cashless than that of B. E has got the maximum number of flats. A number of flats with A is equal to the sum of the number of flats with B and C. Amount of cash with B is double of that with D, but D has got assets worth Rs. 20 lakhs more than that of B.

For how many brothers is it possible to find out the exact value of their total assets?

Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 3
E has got the maximum number of flats and number of flats with A is equal to the sum of a number of flats with B and C. As there are 9 flats, so possible solutions are when E, A, B, C and D have 3, 2, 1, 1 and 2 or 4, 2, 1. 1 and 1 flats respectively.

Case I:B and D have one flat and two flats respectively.

Let the value of a flat be Rs. x. So, the amount of cash with B = x.

So total assets of B = 2x

Total assets of As D has got assets worth 20 lakhs more than that of B

Case II:

B and D have one flat each.

Let value of a flat be Rs. x. So, amount of cash with B = x

⇒ Total assets of B = 2x

Total assets of

As D has got assets worth 20 lakhs more than B

Here the value of x is coming as negative. So this case is not possible.

We can collate the final distribution of assets in the following table:

According to the table B and D buy 1 and 2 flat respectively.

So, Only for B and D, it is possible to find the exact value of their total assets.​

Only 2 brothers find out the exact value of their total assets.

Hence the correct option is B.

IPMAT Mock Test - 8 (New Pattern) - Question 4
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All natural numbers that give remainders 1 and 2 when divided by 6 and 5, respectively, are written in ascending order, side by side from left to right. What is the 99th digit from the left?
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 4
Natural numbers which when divided by 5 give a remainder 2 are given by 5K + 2 [‘k’ is a whole number].

Natural numbers which when divided by 6 give a remainder 1 are given by 6m +1 [‘m’ is a whole number].

5k + 2 = 6m + 1 or 6m - 5k = 1.

All such natural numbers are 7, 37, 67, 97,127

Numbers when written one beside the other

7, 37, 67, 97, 127, 157, 187, 217, 247, 277, 307, 337.......so on.

Observe the series.

The first 4 numbers give first 7 digits. After these 4 numbers, 3 digit numbers follow. The 3 digit numbers from an A.P. with common difference 30.

First three digit number = 127

The 30th 3-digit number is = 127 + (30 -1) × 30

= 997

=> The first 34 numbers (= 4 + 30) give 97 digits (= 7 + 3 × 30)

The next number in this A.P is 997 + 30 = 1027

=> the 99th digit is 0.

Hence answer is an option (d).

IPMAT Mock Test - 8 (New Pattern) - Question 5
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A combo pack having a bulb and a tube light costs Rs. 52. If the cost of the bulb drops by 20% and the cost of the tube light escalates by 50%. the pack would cost Rs. 50. Find the price of a tube light?
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 5
cost of a tubelight = t and cost of a bulb = b.

t + b = 52

cost of bulb drops by 20% and cost of tubelight increases by 50%

⇒1.5t + 0.8b = 50

Solving the two equations in t and b

t + b = 52.......(i)

1.5t + 0.8b = 50.......(ii)

Multiplying (i) equation by 4 and (ii) equation by 5

4t + 4b = 2087.5t + 4b = 250

Subtracting

3.5 t = 42

⇒ t = 12

Thus, the cost of a tube light = Rs. 12

IPMAT Mock Test - 8 (New Pattern) - Question 6
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How many sets of three distinct factors of the number N = 26 x 34 x 52 can be made such that the factors in each set have a highest common factor of 1 with respect to every other factor in that set?
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 6
N = 26 × 34 × 52

Case: [A]

when none of the elements is 1 ⇒ the three factors must be some powers of 2, 3, 5 respectively.

⇒ total number of such sets = 6 × 4 × 2 = 48

Case: [B]

when one of the elements is 1

⇒ the other two factors could be of the form (2a), (3b) − number of sets = 6 × 4 = 24

(26), (52) − number of sets = 6 × 2 = 12

(34),(52) − number of sets = 4 × 2 = 8

(26 × 34 × 52) − number of sets = 6 × 4 × 2 = 48

(26 × 52 × 34) − number of sets = 6 × 2 × 4 = 48

(34 × 52 × 26) − number of sets = 4 × 2 × 6 = 48

∴ total number of such sets = 188

combining both the cases, total sets possible

= 188 + 48 = 236

IPMAT Mock Test - 8 (New Pattern) - Question 7
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Find all the values of p, such that 6 lies some where between the roots of the equation

x2 + 2(p - 3)x + 9 = 0. (‘x’ is a real number).
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 7

Both the roots will be real as 6 lies between the roots

(i) b2 − 4ac > 0

= 4(p − 3)2 − 4 × 9 > 0P(P − 6) > 0

p > 6 or p < />

(ii), since x = 6 lies between the two roots.

so, (6)2 + 2(p − 3)6 + 9 < />

Hence the correct option is A.

IPMAT Mock Test - 8 (New Pattern) - Question 8
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Sum to n terms of the series
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 8

Hence the correct option is D.

IPMAT Mock Test - 8 (New Pattern) - Question 9
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A manufacturing company conducted workshops on three techniques namely designing, formatting and testing for its executives. The number of executives who attended the workshop on designing is twice and thrice of the number of executives who attended the workshops on formatting and testing respectively. 42 executives attended the workshop on formatting. The aggregate number of executives who attended the workshops on exactly two techniques is at most 25% of the aggregate number of executives who attended the workshops on exactly one technique. Workshops on any combination of exactly two techniques were attended by more than 1 executive.

If the minimum possible executives attended the workshop on all three techniques, then find the minimum possible number of executives who attended the workshop on designing only.
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 9
Minimum possible number of executives who attended the workshop on all three techniques = 0.

Since we have to minimise the number of executives who attended the workshop on only designing, we have to maximise the number of executives who attended the workshop on designing, formatting and testing.

A minimum number of employees who attended the workshop on formatting and testing = 2.

A + 2B = 154.

Maximum possible value of B = 25 (Since maximum possible value of B = 0.25A)

The minimum possible value of the number of executives who attended the workshop on only designing.

= 84-(25 - 2) = 61.

IPMAT Mock Test - 8 (New Pattern) - Question 10
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How many integers “a” are there such that 9x2 + 3ax + (a + 5) > 0 for all values of x ?
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 10
If 9x2 + 3ax + (a + 5) > 0 for all values of x, then the discriminant of this quadratic expression must be negative.

⇒ (3a)2 -4(9) (a + 5) < 0 or a2 - 4a - 20 < 0 or R (a - 2)2 < 24

⇒ -2.89

⇒ a = -2, -1,0, 1, 2, 3, 4, 5, 6

⇒ there are 9 such integral values.

Hence (c) is the correct option.

IPMAT Mock Test - 8 (New Pattern) - Question 11
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a, b and c are positive real numbers such that a + b + c = p, where 'p' is a constant. Find the maximum value of (p - a) (p - b)(p - c).
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 11
(p − a), (p − b) and (p − c) all are positive numbers.

∴ AM ≥ GM

Hence (b) is the correct answer.

Alternative Method:

(p − a) + (p − b) + (p − c) = 2p(constant)

⇒ (p − a)⋅(p − b)⋅(p − c) will be maximum when

p − a = p − b = p − c = 2p/3

Maximum value of the product

IPMAT Mock Test - 8 (New Pattern) - Question 12
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What is the remainder when 36001 + 76000 is divided by 13?
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 12
36001 = 3 ⋅ (36000) = 3 ⋅ (36)1000 = 3 ⋅ (729)1000 = 3(728 + 1)1000

728 is divisible by 13, so the remainder is 3.

76000 = (74)1500 = (2401)1500 = (2392 + 9)1500

2392 is divisible by 13, so we have to find the remainder when 91500 is divided by 13.

91500 = (36)500 = (729)500 = (728 + 1)500

since 728 is divisible by 13, therefore the remainder is 1.

So the net remainder = 3 + 1 = 4

IPMAT Mock Test - 8 (New Pattern) - Question 13
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In a shoe store, there are 12 highly expensive pair of shoes. One night, thieves break in and steal 4 shoes, not necessarily having a matching pair among them. What is the probability that at least one matching pair of shoes was stolen?
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 13
Total number of ways to choose 4 shoes

Total number of ways in which no pair is selected

∴ Probability of not getting a pair

Hence, probability of selecting one pair

Hence the correct option is B.

IPMAT Mock Test - 8 (New Pattern) - Question 14
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Read the following graph carefully and answer the question given below.

The bar chart given below shows the import of soya bean (in quintals) for the year 2011 to 2016.

The import of Soya bean in the year 2015 is what per cent of the import of Soya bean in the year 2012?
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 14
Required per cent of the import of Soya bean in the year 2012 = 390 × 100/340 = 114.7

Hence the correct option is C.

IPMAT Mock Test - 8 (New Pattern) - Question 15
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Directions: Read the following graph carefully and answer the question given below.

The bar graph given below shows the per acre yield (in kg) of different countries. Study the graph carefully and answer the questions

By how much percentage is India's per acre yield more than that of Bhutan's?
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 15
India's per acre yield = 200kg Bhutan's per acre yield = 80 kg

Difference in yield = 200 − 80 = 120kg

∴ Required % = 150%

Hence the correct option is B.

IPMAT Mock Test - 8 (New Pattern) - Question 16
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Find out the ratio of milk to that of water, if a mixture is formed after mixing milk and water of 3 same vessels containing the mixture of milk to that of water in the proportion 3:2, 7:3, 9:2 respectively.
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 16
Required mixture

Hence the correct option is C.

IPMAT Mock Test - 8 (New Pattern) - Question 17
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A chemist has 20 litres of a solution that is 20% hydrochloric acid by volume. He wants to dilute the solution to 4% strength by adding water how many litres of water must be added?
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 17
Initial Quantity of Acid Litr And that of water = 16 litre percent of water

Hence the correct option is C.

IPMAT Mock Test - 8 (New Pattern) - Question 18
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The surface area of a sphere is 5024 m2. What is the volume of the sphere? (Approx)
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 18
Surface area of sphere = 4πr2 = 5024.

So r2 = 5024/4π = 400.

So r = 20 m.

Volume of sphere = 33523.8 m3 = 33524 m3 (approx.).

Hence the correct option is A.

IPMAT Mock Test - 8 (New Pattern) - Question 19
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What is the diameter (in cm) of a sphere of surface area 154 sq cm?
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 19
Surface area of sphere 4πr2 = 154cm2

diameter = 2r

diameter = 7cm

Hence the correct option is D.

IPMAT Mock Test - 8 (New Pattern) - Question 20
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A boat running downstream covers a distance of 18 km in 2 hours while for covering the same distance upstream, it takes 3 hours. What is the speed of the boat in still water?
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 20
Speed of boat in downstream = 18 ÷ 2 = 9 kmph

Speed of boat in upstream = 18 ÷ 3 = 6 kmph

Speed of boat in still water = 1 ÷ 2 (9 + 6) = 7.5 kmph

Hence the correct option is D.

IPMAT Mock Test - 8 (New Pattern) - Question 21
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Answer the questions on the basis of the information given below:

A, B, C. D and E are five brothers. They have assets in the form of flats and cash only. Together they have 9 flats and each brother has at least 1 flat. Value (in terms of rupees) of all flats is equal and constant. Cash with each brother is an integral multiple of Rs. 10 lakhs. The amount of cash with B is just enough to buy one more flat, all the other brothers have cashless than that of B. E has got the maximum number of flats. A number of flats with A is equal to the sum of the number of flats with B and C. Amount of cash with B is double of that with D, but D has got assets worth Rs. 20 lakhs more than that of B.

What can be the maximum possible value of assets of A?
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 21
E has got the maximum number of flats and number of flats with A is equal to the sum of a number of flats with B and C. As there are 9 flats, so possible solutions are when E, A, B, C and D have 3, 2, 1, 1 and 2 or 4, 2, 1. 1 and 1 flats respectively.

Case I:B and D have one flat and two flats respectively.

Let the value of a flat be Rs. x. So, the amount of cash with B = x.

So total assets of B = 2x

Total assets of As D has got assets worth 20 lakhs more than that of B

Case II:

B and D have one flat each.

Let value of a flat be Rs. x. So, amount of cash with B = x

⇒ Total assets of B = 2x

Total assets of

As D has got assets worth 20 lakhs more than B

Here the value of x is coming as negative. So this case is not possible.

We can collate the final distribution of assets in the following table:

According to table Maximum possible value of assets of A can be Rs. 110 lakhs.

IPMAT Mock Test - 8 (New Pattern) - Question 22
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Answer the questions on the basis of the information given below:

A, B, C. D and E are five brothers. They have assets in the form of flats and cash only. Together they have 9 flats and each brother has at least 1 flat. Value (in terms of rupees) of all flats is equal and constant. Cash with each brother is an integral multiple of Rs. 10 lakhs. The amount of cash with B is just enough to buy one more flat, all the other brothers have cashless than that of B. E has got the maximum number of flats. A number of flats with A is equal to the sum of the number of flats with B and C. Amount of cash with B is double of that with D, but D has got assets worth Rs. 20 lakhs more than that of B.

What can be the maximum possible value of the assets of all the brothers taken together?
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 22
E has got the maximum number of flats and number of flats with A is equal to the sum of a number of flats with B and C. As there are 9 flats, so possible solutions are when E, A, B, C and D have 3, 2, 1, 1 and 2 or 4, 2, 1. 1 and 1 flats respectively.

Case I:B and D have one flat and two flats respectively.

Let the value of a flat be Rs. x. So, the amount of cash with B = x.

So total assets of B = 2x

Total assets of As D has got assets worth 20 lakhs more than that of B

Case II:

B and D have one flat each.

Let value of a flat be Rs. x. So, amount of cash with B = x

⇒ Total assets of B = 2x

Total assets of

As D has got assets worth 20 lakhs more than B

Here the value of x is coming as negative. So this case is not possible.

We can collate the final distribution of assets in the following table:

As the above table maximum possible value of the assets of all the brothers taken together is:

= Maximum possible value of assets of all the brothers taken together

= 110 + 80 + 70 + 100 + 150.

= 510 lakhs = Rs. 5.1 crore.

IPMAT Mock Test - 8 (New Pattern) - Question 23
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Bag A contains 6 white balls and 4 black balls and bag B contains 3 white balls and 2 black balls. A white ball is picked from bag A and put into bag B. Then, three balls are picked from bag B and put into bag A. Find the probability that a ball picked now from bag A is black.
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 23
The following cases are possible.

I. Three white balls are transferred from bag B to bag A.

Required probability

II. Two white balls and one black ball is transferred from bag B to bag A.

Required probability

III. One white ball and 2 black balls are transferred from bag B to bag A.

Required probability

Required probability that the ball now picked from bag A is black

IPMAT Mock Test - 8 (New Pattern) - Question 24
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Find the area of bounded regions generated by the curves |x| = 3, y = |x| and y = -|x|.
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 24
Given curves can be drawn as

The closed area generated by the three curves is the sum of the areas of the triangles OAB and OCD.

IPMAT Mock Test - 8 (New Pattern) - Question 25
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Three numbers first are twice the second and second is twice the third. The average of the three numbers is 21. Find the largest number of the three.
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 25
Let,

IIIrd number = x

IInd number = 2x

Ist number = 4x

The average of the three numbers is 21.

(x + 2x + 4x)/3 = 21.

7x = 63.

x = 9

Largest number = 4x.

= 4 × 9 = 36.

IPMAT Mock Test - 8 (New Pattern) - Question 26
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Pawan was designing a Mock test. He had 8 questions for this Mock test and he had to assign a total of 30 marks to the 8 questions in this Mock test. If the minimum marks assigned to a question is 2 and each question carries integral marks, then how many ways are possible for Pawan to distribute 30 marks in this Mock test?
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 26
Any question can have a maximum of 16 marks

= q1 + q2 + q3 +……… q8 = 30

where range of marks for any question, say qi

2 ≤ qi ≤ 16

or 0 ≤ q1 − 2 ≤ 14

or 0 ≤ M1 ≤ 14 (where ,qi − 2 = Mi)

⇒ M1 + M2 + M3 +…… M8 = 14

∴ Number of non-negative integral solution for (i)

= 14 + 8 − 1C8−1 = 21C7

Hence the correct option is A.

IPMAT Mock Test - 8 (New Pattern) - Question 27
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Let M be a three-digit number denoted by ‘ABC’ where A. B and C are numerals from 0 to 9. Let N be a number formed by reversing the digits of M. It is known that M - N + 396C is equal to 990. How many possible values of M are there which are greater than 300?
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 27
99A – 99C + 396C = 990

99A + 297C = 990

A + 3C = 10

C = 1, 2, 3 and A = 7, 4, 1.

For each value of A and C, B can take 10 values. Also, A has to be greater than or equal to 3. So, 4B2 and 7B1 will be the numbers. B can take any value from 0 to 9. So, 20 values in total.

IPMAT Mock Test - 8 (New Pattern) - Question 28
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A manufacturing company conducted workshops on three techniques namely designing, formatting and testing for its executives. The number of executives who attended the workshop on designing is twice and thrice of the number of executives who attended the workshops on formatting and testing respectively. 42 executives attended the workshop on formatting. The aggregate number of executives who attended the workshops on exactly two techniques is at most 25% of the aggregate number of executives who attended the workshops on exactly one technique. Workshops on any combination of exactly two techniques were attended by more than 1 executive.

If the maximum possible executives attended the workshop on all three techniques, then find the maximum possible number of executives who attended the workshop on both designing and formatting but not testing.
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 28
Maximum possible number of executives who attended the training sessions on all the three techniques.

= 28 - (x3 + xs + x6)

= 28 - (0 + 2 + 2) = 24.

Now,

A + 2B = 154

⇒ A + 2B = 154 - 72 = 82.

Since the maximum value of B can be 0.25A, therefore the maximum possible value of B can be 13.

Therefore the maximum possible number of executives who attended the workshop on designing and formatting.

= 13 - (2 + 2) = 9.

IPMAT Mock Test - 8 (New Pattern) - Question 29
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25x2 + y2 = 34 and xy = 3, then value of (125x3 + y3) is:
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 29
Given: 25x2 + y2 = 34

We know that,

(x3 + y3) = (x + y) (x2 + y2 − xy)

(x + y)2 = x2 + y2 + 2xy

Now,

25x2 + y2 = 34

⇒ 25x2 + y2 + 10xy = 34 + 10xy

⇒ 25x2 + y2 + 10xy = 34 + 30

⇒ (5x + y)2 = 64

⇒ (5x + y) = 8

So,

(125x3 + y3)

= (5x + y)(25x2 + y2 − 5xy)

= (8)(34 − 15)

= 8 × 19

= 152

∴ The value of (125x3 + y3) is 152.

Hence, the correct option is (A).

IPMAT Mock Test - 8 (New Pattern) - Question 30
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A natural number is randomly selected from 100 to 360. Find the probability that it is divisible by exactly two of the three numbers 4, 6 and 8.
Detailed Solution for IPMAT Mock Test - 8 (New Pattern) - Question 30
Total number of numbers = 360 − 100 + 1 = 261

LCM of 4,6 and 8 is equal to 24

Here, we will take 3 cases.

Case 1: Numbers divisible by 4 and 6 but not 8.

Case 2: Numbers divisible by 6 and 8 but not 4.

Case 3: Numbers divisible by 8 and 4 but not 6.

Case 1: Numbers of the form 12, (12 + 24),(12 + 2 × 24),…

From 100 to 360, the numbers are from (12 + 4 × 24) to (12 + 14 × 24) = (14 − 4 + 1) numbers = 11 numbers.

case 2:

Numbers divisible by 8 have to be divisible by 4

Hence, no number exists in this case.

case 3:

Numbers of the form 8, (8 + 24), (8 + 2 × 24),… and

16, (16 + 24), (16 + 2 × 24),…

From 100 to 360 , the numbers are from (8 + 4 × 24) to (8 + 14 × 24) and (16 + 4 × 24) to (16 + 14 × 24)

= (14 − 4 + 1) + (14 − 4 + 1) = 22

Required probability

Alternative Method:

LCM of 4,6 and 8 is equal to 24

Number of natural numbers that are divisible by 24

Number of natural numbers that are divisible by 8

= (45 − 12 − 11) = 22

Number of natural numbers that are divisible by 4

= (30 − 8 − 11) = 11

Required probability

Hence the correct option is A.

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IPMAT Mock Test - 8 (New Pattern) MCQs with Answers

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Online Tests for IPMAT Mock Test - 8 (New Pattern) IPMAT Mock Test Series

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