UIET Placement Test: December 29, 2020


40 Questions MCQ Test EduRev Placement Drive 2020 | UIET Placement Test: December 29, 2020


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This mock test of UIET Placement Test: December 29, 2020 for Computer Science Engineering (CSE) helps you for every Computer Science Engineering (CSE) entrance exam. This contains 40 Multiple Choice Questions for Computer Science Engineering (CSE) UIET Placement Test: December 29, 2020 (mcq) to study with solutions a complete question bank. The solved questions answers in this UIET Placement Test: December 29, 2020 quiz give you a good mix of easy questions and tough questions. Computer Science Engineering (CSE) students definitely take this UIET Placement Test: December 29, 2020 exercise for a better result in the exam. You can find other UIET Placement Test: December 29, 2020 extra questions, long questions & short questions for Computer Science Engineering (CSE) on EduRev as well by searching above.
QUESTION: 1

If 30 females from each park are above 80 years age then find the average of no. of females who are below or equal to the age of 80 years from all the parks. 

Solution:

Total female population = 150 + 200 + 350 + 450 + 500 = 1650 

Female population above 80 years age = 30 × 5 = 150 

Required average = 1650−150/ 5 = 300

QUESTION: 2

Manoj gave 60% of his salary to his wife and invested the rest amount in mutual funds. His wife spends 30% amount on grocery and 20% on rent. From the remaining amount, she purchased gold worth Rs. 18000. Find salary of Manoj. 

Solution:

let salary of Manoj be Rs 100x

The amount given to wife = 60/100 × 100x = Rs. 60x

ATQ, 60x × 50/100 = 18000

x = 600

Salary of Manoj = 100x = Rs 60000

QUESTION: 3

In the following two equations questions numbered (I) and (II) are given. You have to solve both equations and Give answer 

I. 2x2 - 7x – 60 = 0 
II. 3y2 + 13y + 4 = 0

Solution:

I. 2x2 - 7x – 60 = 0 

2x2 - 15x + 8x – 60 = 0 

x (2x – 15 ) + 4 (2x – 15) = 0 (x + 4) (2x − 15) = 0 

x = −4, 15/2 

II. 3y2 + 13y + 4 = 0 

3y2 + 12y + y + 4 = 0 

3y (y + 4) + 1 (y + 4) = 0 

(3y + 1) (y + 4) = 0 y = − 1/3 , −4 

Hence, No relation between x and y

QUESTION: 4

A shopkeeper marked the price of an article by 40% above cost price and gave discount of Rs. 224. On the final amount, he charged 10% tax. In the whole transaction, he earned Rs. 158.6. Find cost price of the article. 

Solution:

let cost price be Rs. 100x 

Marked price = 140 /100 × 100x = Rs 140x 

Selling price = Rs (140x − 224) 

Selling price after tax = 110/100 × (140x − 224) = Rs (154x − 246.4) 

ATQ, 100x + 158.6 = 154x − 246.4 

x = 7.5 

Cost price of article = 100x = Rs 750

QUESTION: 5

A club has 256 members of whom 144 can play football, 123 can play tennis, and 132 can play cricket. Moreover, 58 members can play both football and tennis, 25 can play both cricket and tennis, while 63 can play both football and cricket. If every member can play at least one game, then the number of members who can play only tennis is

Solution:

From observing the data given, we find that it is a closed 3 set Venn diagram.

Let the three sports be F, T and C for Football, Tennis and Cricket respectively
n (F U T U C) = 256 , n(F) = 144, n(T) = 123, n(C) = 132, n(F ∩ T) = 58, n(C ∩T) = 25, n(F ∩ C) = 63

We know that (A U B U C) = n(A) + n(B) +n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩B ∩ C)
So, 256 = 144 + 123 + 132 - 58 - 25 - 63 + n (F ∩ T ∩ C)
n (F ∩ T ∩ C) = 256 - 144 + 123 +132 - 146
n (F ∩ T ∩ C) = 256 - 253 = 3
Now, it is easy to calculate the number of students who only play tennis using a Venn diagram.
n (Students who play only Tennis) = 123 - (55 + 3 + 22) = 123 - 80
n (Students who play only Tennis) = 43 students

QUESTION: 6

Meena scores 40% in an examination and after review, even though her score is increased by 50%, she fails by 35 marks. If her post-review score is increased by 20%, she will have 7 marks more than the passing score. The percentage score needed for passing the examination is

Solution:

Meena scores 40 % in an exam

After review, she scores 50 % more => Increase of 50 % from 40 % = 40% + 20% = 60% She fails by 35 marks, by scoring 60%

60% score = Pass mark - 35 ----- (1)

If her post review score is increased by 20%, she would have 7 more than the pass mark. 20% of 60% = 12 %

So, 60% + 12% = 72% of marks = Pass mark + 7 ------ (2)

So, 12% marks = 35 + 7 (5: 1 ratio)

So, similarly 12% can be re written as 10 % and 2 % (maintaining the 5:1 ratio)

Hence the pass percentage = 60 % + 10 % = 70%

(or)

Pass percentage = 72 % - 2 % = 70%

QUESTION: 7

Sum of three Whole numbers a, b and c is 10. How many ordered triplets (a, b, c) exist?

Solution:

a + b + c = 10. a, b, c are whole numbers. Now this is similar to the previous question that we solved by placing 10 sticks and simplifying.

We cannot follow an exactly similar approach, as in this case a, b and c can be zero. Let us modify the approach a little bit. Let us see if we can remove the constraint that a, b, c can be zero.

If we give a minimum of 1 to a, b, c then the original approach can be used. And then we can finally remove 1 from each of a, b, c. So, let us distribute 13 sticks across a, b and c and finally remove one from each.

a + b + c = 13. Now, let us place ten sticks in a row

|       |       |       |       |       |       |       |       |       |       |       |       |

This question now becomes the equivalent of placing two '+' symbols somewhere between these sticks. For instance,

|       |       |       |  +  |       |       |       |       |  +  |       |       |       |,

This would be the equivalent of 4 + 5 + 4. or, a = 4, b = 5, c = 4.

There are 12 slots between the sticks, out of which one has to select 2 for placing the '+'s. The number of ways of doing this is 12C2. Hence, correct answer is 66.

QUESTION: 8

a, b, c are three distinct integers from 2 to 10 (both inclusive). Exactly one of ab, bc and ca is odd. abc is a multiple of 4. The arithmetic mean of a and b is an integer and so is the arithmetic mean of a, b and c. How many such triplets are possible (unordered triplets)?

Solution:

 Exactly one of ab, bc and ca is odd
=> Two are odd and one is even.

abc is a multiple of 4
=> the even number is a multiple of 4.

The arithmetic mean of a and b is an integer
=> a and b are odd.

and so is the arithmetic mean of a, b and c.
=> a + b + c is a multiple of 3.

c can be 4 or 8.

c = 4; a, b can be 3, 5 or 5, 9

c = 8; a, b can be 3, 7 or 7, 9

Four triplets are possible.

The question is "How many such triplets are possible (unordered triplets)?"

Hence the answer is "4"

Choice D is the correct answer.

QUESTION: 9

Let x and y be positive real numbers such that log5(x + y) + log5(x - y) = 3, and log2y - log2x = 1 - log23. Then xy equals

Solution:

Given that, log5(x + y) + log5(x - y) = 3
We know log A + log B = log (A x B)
log5(x + y) + log5(x - y) = log5(x2 - y2)
log5(x2 - y2) = 3
x2 - y= 53
x2 - y= 125
Similarly, log2y - log2x = 1 - log23
log2 y/x = log22 - log23
logy/x = log2 2/3
3y = 2x
(3/2 y)2 - y2 = 125
9/2 y2 - y2 = 125
5/4 y2 = 125
y2 = 100
y = 10
x = 15
So, xy = 15 × 10 = 150

QUESTION: 10

A man invested a certain sum in scheme A at 15% p.a. for 2 years and earned Rs 1950 as simple interest. He increased his sum by Rs. ‘x’ and invested in another scheme B at 10% p.a. C.I. for 2 years and received Rs. 1680 as compound interest. Find the value of ‘x’ ? 

Solution:

Sum = (1950 × 100) /(2 × 15) =Rs 6500 

CI in 2 years at 10% per annum= 10 + 10 + (10 × 10)/ 100 = 21% 

ATQ (6500 + x) × 21/100 = 1680
⇒ (6500 + x) = 8000 x = Rs 1500

QUESTION: 11

In a certain exam, the ratio of the number of students passed to the number of students failed is 7: 4. Has 8 more students passed, the ratio of the number of students passed to the number of students failed would have been 9: 2. Find the number of students who failed in the exam.

Solution:

QUESTION: 12

Ram bought a bike at 20% discount on MRP. After 1 year Ram sell the bike to Ramesh at 10% loss. After 1 year more Ramesh sell the bike at 20% profit to Ranjan. If Ranjan paid Rs. 1,29,600, then find the M.R.P. of the bike ? 

Solution:

Ram’s cost price = M.R. P.× 80/100 

Ramesh C. P. = M. R. P.× 80/100 × 90/100 

Ranjan C. P. = M.R. P.× 80/100 × 90/100 × 120/100 = 1,29,600 

⇒ M.R.P. = Rs. 1,50,000

QUESTION: 13

Directions: Read the data carefully and answer the questions.

There are 450 coupons which can be used in Pedicure and Hair cutting. Ratio between Males to Females who use their coupons in Hair cutting is 13 : 7 Number of males who use their coupons in Pedicure is 72 more than number of females who use their coupon in Hair cutting. Total number of males who use their coupon in Pedicure and Haircutting together is 174 more than total number of females who use their coupon in Pedicure and Haircutting together. 

Males who use their coupon in Pedicure is what percent of the Males who use their coupons in Haircutting? 

Solution:

Let, Males and females who use their coupons in Haircutting be 13x and 7x respectively.
⇒ Males who use their coupons in Pedicure = 7x + 72
Then Females who use their coupons in Pedicure = 450 − 13x − 7x − 7x − 72 = 378 − 27x 

ATQ,
7x + 72 + 13x − (7x + 378 − 27x) = 174
40x − 306 = 174
40x = 480 x = 12 

QUESTION: 14

Directions: Read the data carefully and answer the questions.

There are 450 coupons which can be used in Pedicure and Hair cutting. Ratio between Males to Females who use their coupons in Hair cutting is 13 : 7 Number of males who use their coupons in Pedicure is 72 more than number of females who use their coupon in Hair cutting. Total number of males who use their coupon in Pedicure and Haircutting together is 174 more than total number of females who use their coupon in Pedicure and Haircutting together. 

Find the ratio between Total number persons who use their coupons in Pedicure to total number of persons who use their coupons in Haircutting? 

Solution:

Required Ratio = 156 + 54/156 + 84 

= 210/240 = 7/ 8

QUESTION: 15

Directions: Read the data carefully and answer the questions.

There are 450 coupons which can be used in Pedicure and Hair cutting. Ratio between Males to Females who use their coupons in Hair cutting is 13 : 7 Number of males who use their coupons in Pedicure is 72 more than number of females who use their coupon in Hair cutting. Total number of males who use their coupon in Pedicure and Haircutting together is 174 more than total number of females who use their coupon in Pedicure and Haircutting together. 

Ratio between Males who use their coupon in Pedicure to that of in Spa is 4 : 5, while ratio between Females who use their coupon in Haircutting to that of in Spa is 6 : 11. Find total number of people who use their coupons in Spa?

Solution:

Males who use their coupons in Spa = 156 × 5 /4 = 195 

Females who use their coupons in Spa = 84 × 11 /6 = 154 

Total number of people who use their coupon in Spa = 195 + 154 = 349

QUESTION: 16

In a box there are 6 blue ball, X red balls & 10 green balls. Probability of choosing one red ball from the given box is 1/3, then find the sum of red and blue balls in the box? 

Solution:

ATQ, 

x = 8

∴ Sum of red and blue ball = 8 + 6 = 14

QUESTION: 17

What is the probability of forming word from the letters of word “IMPEACH” such that all vowels come together? 

Solution:

Total numbers of ways → 7! 

Favorable numbers of ways → 5! ×3! 

Probability → =1/7

QUESTION: 18

If the difference between the simple interest and compound interest on some principal amount at 20% per annum for 3 years is ` Rs. 48, then the principle amount must be 

Solution:

Solve using options. If we try 500 (option b) for convenience, we can see that the difference between the two is 64 (as the SI would amount to 300 and CI would amount to 100 + 120 + 144 = 364).
Since, we need a difference of only 48 we can realize that the value should be 3/4th of 500. Hence, 375 is correct

QUESTION: 19

If Ajit saves Rs. 400 more each year than he did the year before and if he saves Rs. 2000 in the first year, after how many years will his savings be more than Rs.100000 altogether? 

Solution:

We need the sum of the series 2000 + 2400 + 2800 to cross 100000.
Trying out the options, we can see that in 20 years the sum of his savings would be: 2000 + 2400 + 2800 +…+ 9600.
The sum of the series would be 20 × 5800 = 116000.
If we remove the 20th year we will get the saving for 19 years.
The series would be 2000 + 2400 + 2800 + … + 9200. Sum of the series would be 116000 − 9600 = 106400.
If we remove the 19th year’s savings the savings would be 106400 − 9200 which would go below 100000.
Thus, after 19 years his savings would cross 100000. Option (a) is correct

QUESTION: 20

The average rainfall for Monday, Tuesday and Wednesday is 4.3 cm. The average rainfall for Friday, Saturday and Sunday is 3.9 cm. If the average rainfall for the total week is 3.7 cm, then what is the rainfall recorded on Thursday? 

Solution:

Total rainfall for the week = 3.7 × 7 = 25.9 cm ...(i)
Rainfall for Monday, Tuesday and Wednesday = 4.3 × 3 = 12.9 cm ...(ii)
Rainfall for Friday, Saturday and Sunday = 3.9 × 3 = 11.7 cm ...(iii)
Rainfall for Thursday = Equation (i) – Equation (ii) – Equation (iii) = 25.9– 12.9 – 11.7 = 1.3 cm

QUESTION: 21

The average weight of 23boxes is 3kg. If the weight of the container (in which the boxes are kept) is included, the calculated average weight per box increases by 1 kg. What is the weight of the container? 

Solution:

The average weight per box is asked.
Hence, the container does not have to be counted as the 5th item.
Also, since the average for 23 boxes goes up by 1 kg, the total weight must have gone up by 23 kgs.
That weight is the actual weight of the container.
Hence, option (d) is correct. 

QUESTION: 22

On March 1st 2016, Sherry saved ₹1. Everyday starting from March 2nd 2016, he saved ₹1 more than the previous day. Find the first date after March 1st 2016 at the end of which his total savings will be a perfect square. 

Solution:

n(n + 1)/2 should be a perfect square. The first value of n when this occurs would be for n = 8.
Thus, on the 8th of March the required condition would come true

QUESTION: 23

The odds in favour of standing first of three students Amit, Vikas and Vivek appearing at an examination are 1 : 2. 2 : 5 and 1 : 7 respectively. What is the probability that either of them will stand first (assume that a tie for the first place is not possible). 

Solution:

P (Amit) = 1/3 P (vikas) = 2/7 P (vivek) = 1/8. Required Probability = 1/3 + 2/7 + 1/8 = 125/168.

QUESTION: 24

A car started its journey with its usual speed but after travelling for 5 hours the car meet with an accident so the speed of the car reduced by 8% and it took 6 hours to cover the remaining 276 km then find the percentage by which distance travelled by the car in first 5 hours is less than the remaining distance covered by the car? 

Solution:

Speed of the car in covering the remaining distance= 276/6= 46 kmph
Original speed of the car= 46*100/92= 50kmph
Distance travelled by the car in first 5 hours= 50*5= 250 km
Difference between distance travelled= 276-20= 26
Required percentage= 26*100/276= 9.42%

QUESTION: 25

There are three persons A, B and C. A invested 20% more than that of B and C invested 40% more than that of B. If the difference between the investment of A and C is Rs 4000 than find the investment of A?

Solution:

Let the let the money invested by B be x Money invested by A= x*120/100= 6x/5 Money invested by C= x*140/100= 7x/5 So, 7x/5 – 6x/5= 4000 x/ 5=4000 x= Rs 20000

So, Money invested by A= 20000*120/100= Rs 24000

QUESTION: 26

In a township of 1000 families, some families have an SUV and some families have a car. Some families have both the vehicles and some families have no vehicle. It is known that 350 families own an SUV. Out of those who own an SUV, 50% own a car too. If it is known that 75% of the families of the township owns at least one vehicle, how many families own at most one vehicle?

Solution:

No. of families who owns an SUV = 350
No. of families who owns both SUV and car = 50% of 350 = 175
It is given that 75% of the families own at least one vehicle
So, 250 families do not own any vehicle
No. of families who own a car and not an SUV = 750 – 350 = 400
So, no of families who own at most one vehicle
= 250 + (350 – 175) + (750 – 350) = 825
Hence, 825 is the correct answer

QUESTION: 27

A shopkeeper makes a profit on the sale of a material by marking the price 20% more than cost price on a normal day. On a particular day, while buying he gets 10% extra material for a given price. If he gives 25% discount on the marked prices, what is the profit or loss percentage on the transaction? 

Solution:

Assume the cost of material = Re 1/unit and normal marked price = 1.2/unit 

Consider the amount of money spent on the given day = Rs. 100 

Then the shopkeeper will get 110 units of material.
The marked price of 110 units = 110*1.2 

After giving 25% discount, new selling price = 110*1.2*0.75 = 99
Hence on the whole transaction he makes 100 - 99 = 1% loss

QUESTION: 28

A milkman bought 15 litres of milk and mixed it with 3 litres of mineral water (which is not free). He claims to his customers, who do not know about mixing, that he is making a profit of 10% only. However, his actual profit is 20%. What is the ratio of the cost of milk/litre and water/litre. 

Solution:

Let the CP of milk be Rs. 100/litre
As he is claiming 10% profit, SP of the total mixture = Rs. [(15 + 3) * 110] = Rs. 1980
Actual profit = 20%
Let actual CP be Rs. x.
Then, x + 20% of x = Rs. 1980
Solving for x, we get x = Rs. 1650
So, Actual CP = Total CP of milk + Total CP of water
Or, Total CP of water = Rs. 1650 – (15 * 100) = Rs. 150
Or, CP of water = Rs. 50/litre
Required ratio = 100:50 = 2:1
Hence, option C is correct.

QUESTION: 29

A team of miner planned to mine 1800 tonnes in a certain number of days.Due to some difficulties in one third of the planned days, the team was able to achieve an output of 20 tons of ore less than the planned output.To make up for this, the team overachieved for the rest of the days by 20 tons.The end result for this that they completed the one day ahead of time.How many tone of ore did the team initially plan to ore per day?

Solution:

Let us assume the no. of days as ‘3d’ and the output per day as ‘x’

Then, 3d*x = 1800 … (1) 

For the first ‘d’ days, the output was (x - 20).

For the next ‘2d - 1’ days, the output was (x + 20). 

=> d(x-20) + (2d - 1) (x + 20) = 1800

=> dx – 20d + 2dx + 40d – x – 20 = 3dx {Replacing 1800 with 3dx from equation (1)}

=> 20d = x + 20

=> d = (x + 20)/20 … (2) 

=> 3 [(x + 20)/20] x = 1800

=> x2 + 20x = (1800/3) *20

=> x2 + 20x – 12000 = 0

=> (x + 120) (x - 100) = 0

=> x = -120 or 100 

Since x is the output it cannot be negative.

So, the initial planned output is 100 tonnes. Thus, Option B

QUESTION: 30

Two trains, Garibrath express and Durunto express are moving towards each other on parallel tracks. The speed of Garibrath and Durunto express are 72 km/hr and 54km/hr respectively. Ram is sitting near the front end of Garibrath and Shyam is sitting near the rear end of Durunto express. As soon as the trains start crossing each other, Ram starts moving towards the rear end of Garibrath at the speed of 3 m/s and Shyam starts to move towards the front end of Durunto at the speed of 4 m/s. If the lengths of Garibrath and Durunto express are 120 m and 180 m respectively. After how much time(in seconds) from the instant that trains start crossing each other, will Ram and Shyam cross each other?

Solution:

Speed of Garibrath express = 72 km/hr = 20m/s

Speed of Durunto express = 54 km/hr = 15m/s

When the trains start to cross each other, Ram is at front end of Garibrath express and Shyam is at rear end of Durunto express. So the initial distance between them is equal to the length of Durunto express.

Hence initial distance between them = 180m

Shyam is moving in the same direction as the train so his effective speed is 15+4 = 19 m/s

Ram is moving in the direction opposite to the train, so his effective speed is 20-3 =17 m/s.

Hence with reference to the train Ram and Shyam are moving in the same direction but with reference to ground they are moving in the opposite direction with relative speed of 19+17 = 36 m/s

Total distance to be covered = 180 m

Hence required time = 180/36 = 5 seconds 

QUESTION: 31

A student was given 8 two-digit numbers to add,by a teacher.If the student reversed each number and added the results,the sum of the numbers would be 36 more than the actual sum.Find the excess of the sum of the units digits over that of the ten digits.

Solution:

Let us say that the 8 numbers are a1 b1, a2 b2, a3 b3 … a8 b8 …{eg: 38}

They actually represent a1*10 + b1, a2*10 + b2 … a8*10 + b8 …{eg: 3*10 + 8}

Sum of the numbers = (a1 + a2 + a3 .. a8)*10 + (b1 + b2 + b3 .. b8)

Reversed numbers will be b1 a1, b2 a2, … b8 a8

Sum of the reversed numbers will be (b1 + b2 + b3 .. b8)*10 + (a1 + a2 + a3 .. a8)

Reverse sum – actual sum = 36 … {Given in the question}

(b1 + b2 + b3 .. b8)*9 – (a1 + a2 + a3… a8)*9 = 36

=> (b1 + b2 + b3 … b8) – (a1 + a2 + a3 … a8) = 4 {9 can be taken common and cancelled from both sides}

=> Sum of the units digits – Sum of the tens digits = 4

QUESTION: 32

An open box is made of wood 2 cm thick. Its internal length is 86 cm, breadth is 46 cm and height is 38 cm. The cost of painting the outer surface of the box at Rs. 10 per m2 (excluding outer upper edge) is 

Solution:

Internal length = 86 cm
So, external length = (86 + 2 + 2) = 90 cm
Internal breadth = 46 cm
External breadth = (46 + 2 + 2) = 50 cm
Internal height = 38 cm
External height = (38 + 2) = 40 cm
Outer surface area = 2 × 90 × 40 + 2 × 50 × 40 + 90 × 50 = 7200 + 4000 + 4500 = 15700 cm2 = 1.57 m2
Cost of painting the outer surface at Rs. 10 per m2 = 1.57 × 10 = Rs. 15.7

QUESTION: 33

Two jars contain equal quantities of 40% alcohol. Anju changed the concentration of the first jar to 50% by adding extra quantity of pure alcohol. Sonali changed the concentration of the second jar to 50% by replacing a certain quantity of the solution with pure alcohol. By what percentage is the quantity of alcohol added by Anju more than that replaced by Sonali ? 

Solution:

Let initial quantity of solution be 100 ml and therefore, alcohol= 40 ml

Let y ml of alcohol is added to 1st solution

For first jar,

∴ 80 + 2y= 100 + y ⇒ y = 20 ml For second jar, let y ml of solution is replaced with y ml of alcohol

Hence, required percentage

QUESTION: 34

The average age of a group of six children is 15 years. From the group, two children left, whose ages were 3 years more and 5 years more than the average age. 4 new children, whose average age is 4 years more than the given average age, join the group. Find the new average age.

Solution:

Sum of the ages of 6 children = 15 × 6 = 90
When two children left, sum of the ages of 4 children = 90 − (18 + 20) = 52
Sum of the ages of new children = (15 + 4) × 4 = 76
∴ Required average = 76 + 52 /8 = 128 /8 = 16 years

QUESTION: 35

The total number of men, women and children working in a factory is 18. They earn Rs. 4000 in a day. If the sum of the wages of all men, all women and all children is in the ratio of 18 : 10 : 12 and if the wages of an individual man, woman and child is in the ratio 6 : 5 : 3, then how much a women earn in a day ? 

Solution:

Ratio of number of men, women and children = 18 /6 ∶ 10/ 5 : 2/ 3 = 3x ∶ 2x ∶ 4x
⇒ (3x + 2x + 4x) = 18
⇒ x = 2 Now, number of women = 4 
Share of all women = 10 /40 × 4000 = Rs. 1000
∴ Share of each women = 1000 4 = Rs. 250

QUESTION: 36

The sum of three consecutive odd numbers and three consecutive even numbers together is 231. Also, the smallest odd number is 11 less than the smallest even number. What is the sum of the largest odd number and the largest even number? 

Solution:

The smallest odd number = x

∴ The smallest even number = x + 11

∴ x + x + 2 + x + 4 + x + 11 + x + 13 + x + 15 = 231

=> 6x + 45 = 231

=> 6x = 231 - 45 = 186

∴ x = 186/6 = 31

∴ Required sum

= x + 4 + x +15

= 2x + 19 = 2 × 31 + 19

= 62 + 19 = 81

QUESTION: 37

The L.C.M of two number is 3 times of largest number and difference between smallest number and H.C.F. of two number is 12. Then find out the smallest number.

Solution:

Let two number are x and y(x > y)

L.C.M of both number = 3x

l.c.m × H.C.F = x × y

3x × H.C.F = x × y

ysmallest number = 3 × H.C.F

ATQ,

y - H.C.F = 12

3 × H.C.F - H.C.F =12

H.C.F = 6

so, smallest number y= H.C.F +12

= 6+12

= 18

QUESTION: 38

If 33% of the sum of two number is equal to 67% of difference of the same numbers. If sum of same two number equal to 469. Then find out the smaller number?

Solution:

Let two number are A and B.

ATQ,

33% of (A + B) = 67% of (A - B)

34%A = 100%B

A/B = 50/17

Let A = 50x

And B = 17x

ATQ,

50x + 17x = 469

x = 7

Smaller number = 17x = 177

= 119

QUESTION: 39

One -third of certain journey is covered at the rate of 25kmph,one-fourth at the rate of 30kmph and the rest at 50 kmph.Find the average speed for the whole journey.

Solution:

Average speed = total distance / total time

LCM = 3, 4 = 12

Average speed = 12 / (1/3*12*1/25 + ¼ * 12*1/30 +5/50)

= 12/ (4/25 + 3/30+ 5/30

= 150*12/54 = 33.333 kmph

QUESTION: 40

The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B to 9 a.m. and travels towards A at 75 km/hr. At what time do they meet?

Solution:

Distance travelled by first train in one hour = 60 × 1 = 60 km 
Therefore, 
distance between two train at 9 a.m. = 330 – 60 = 270 km

Now,

Relative speed of two trains = 60 + 75 = 135 km/hr 

Time of meeting of two trains = 270/135 = 2 hrs. 

Therefore, both the trains will meet at 9 + 2 = 11 A.M.