Instructions

The following pie chart shows the percentage distribution of runs scored by a batsman in a test innings.

Question 81

If 5 of the dot balls had been hit for 4s, and if two of the shots for which the batsman scored 3 runs each had fetched him one run instead, what would have been the central angle of the sector corresponding to the percentage of runs scored in 4s? {Use data from the previous question}

Video Solution

Solution

Net increment in the number of runs due to 5 the dot balls resulting in 4s and two 3s resulting in two 1s = 4*5 - 2*2 = 16.

Therefore, the total number of runs scored by the batsman = 16 + 306 = 322 runs.

The number of runs scored by 4's = 132 + 20 = 152 {We know that batsman scored 132 runs from 4s alone}

The angle subtended by 4's = $$\dfrac{152}{322}\times 360$$ = 169.93 $$\approx$$ 170.

Instructions

For the following questions answer them individually

Question 82

Carpenter Rajesh has a circular piece of plywood of diameter 30 feet. He has cut out two disks of diameter 20 feet and 10 feet. What is the diameter of the largest disk that can be cut out from the remaining portion of the plywood piece?

Video Solution

Solution

It is given that the 3rd disk also has the maximum diameter. This is possible only when all 3 discs touch each other externally.

Also the sum of the diameter of the two disks is 10+20 = 30 feets. Hence, we can say that centres of these two disk lie on a diameter of original plywood.The figure can be drawn as shown below,

Here, in the diagram O is the centre of original disk. A and C are the centres of disks having radius 10 feet and 5 feet respectively. D is the touching point of circles with centre A and C.

Let 'x' be the radius of the disk that is to be cut out from the remaining part.

We can say that AO = OD = CD = 5 feet. Also, AB = 10+x, BC = 5+X, OB = 15-x and AC = 15 feet.

Also, let 'y' be the length of BD.

By applying apollonius theorem in triangle ABD,

$$AB^2+BD^2=2(OB^2+OD^2)$$

$$(10+x)^2+y^2=2((15-x)^2+5^2)$$

$$x^2+20x+100+y^2=500-60x+2x^2$$

$$y^2 - x^2+80x=400$$ ... (1)

Similarly, by applying apollonius theorem in triangle OBC,

$$OB^2 + BC^2 = 2(BD^2+DC^2)$$

$$(15-x)^2 + (5+x)^2 = 2(y^2+5^2)$$

$$2x^2-20x+250=2y^2+50$$

$$2x^2-2y^2-20x+200=0$$ ... (2)

By equation (1) and (2) we can say that,

$$160x-20x+200=800$$

$$x=600/140 = 30/7$$

Therefore, the diameter of the largest disk that can be cut out from the remaining portion of the plywood piece = 2*30/7 = 8.57 feet. Hence, option C is the correct answer.

Question 83

Lionel and Ronaldo had a discussion on the ages of Jose’s sons. Ronaldo made following statements about Jose’s sons:
i. Jose has three sons.
ii. The sum of the ages of Jose’s sons is 13.
iii. The product of the ages of the sons is the same as the age of Lionel.
iv. Jose’s eldest son, Zizou weighs 32 kilos.
v. The sum of the ages of the younger sons of Jose is 4.
vi. Jose has fathered a twin.
vii. Jose is not the father of a triplet.
viii. The LCM of the ages of Jose’s sons is more than the sum of their ages.
Which of the following combination gives information sufficient to determine the ages of Jose’s sons?

Video Solution

Solution

Statement iii and iv are redundant. Thus, we can cancel out first three options.

From (i) , (ii) , (v), we have :

Let A,B,C be the ages of Jose's sons in ascending order

and $$A + B + C = 13$$

and $$A + B = 4$$

=> $$C = 13 - 4 = 9$$

Now, from (vi) we get $$A = B$$

=> $$A = B = \frac{4}{2} = 2$$

$$\therefore$$ i, ii, v and vi are required to answer the question.

Question 84

Ram and Shyam form a partnership (with Shyam as working partner) and start a business by
investing 4000 and 6000 respectively. The conditions of partnership were as follows:
1. In case of profits till 200,00 per annum, profits would be shared in the ratio of the invested capital.
2.Profits from 200,001 till 400,000 Shyam would take 20% out of the profit, before the division of remaining profits, which will then be based on ratio of invested capital.
3.Profits in excess of 400,000, Shyam would take 35% out of the profits beyond 400,000, before the division of remaining profits, which will then be based on ratio of invested capital.
If Shyam’s share in a particular year was 367000, which option indicates the total business
profit (in ) for that year?

Video Solution

Solution

Ratio of profits earned by Ram : Shyam = 4000 : 6000

= 2 : 3

If profit < 2,00,000

% of profit earned by Shyam = $$\frac{3}{5} \times$$ 100 = 60%

If 2,00,000 < profit < 4,00,000, he gets 20 % and 60 % of the remaining profit.

% of profit earned by Shyam = 20% + .80 $$\times$$ 60% = 68%

If profit > 4,00,000

% of profit earned by Shyam = 35 % + .65 $$\times$$ 60% = 74%

Now, for first 2,00,000 profit earned by Shyam = $$\frac{60}{100} \times$$ 2,00,000 = Rs. 1,20,000

For second 2,00,000 profit earned by Shyam = $$\frac{68}{100} \times$$ 2,00,000 = Rs. 1,36,000

Let total profit earned by them = Rs. (4,00,000 + $$x$$)

=> From $$Rs. x$$ profit, Shyam received = 3,67,000 - 1,20,000 - 1,36,000 = Rs. 1,11,000

=> $$\frac{74}{100} \times x$$ = 1,11,000

=> $$x$$ = 1,11,000 $$\times \frac{100}{74}$$ = 1,50,000

$$\therefore$$ Total profit = 4,00,000 + 1,50,000 = Rs. 5,50,000

Question 85

A property dealer bought a rectangular piece of land at 1000/sq. ft. The length of the plot is less than twice its breadth. Due to its size, there were no buyers for the full plot. Hence he decided to sell it in smaller sized pieces as given below. The largest square from one end was sold at 1200/sq. ft. From the remaining rectangle the largest square was sold at 1150/sq. ft. Due to crash in the property prices, the dealer found it difficult to make profit from the sale of the remaining part of the land. If the ratio of the perimeter of the remaining land to the perimeter of the original land is 3 : 8, at what price (in ) the remaining part of the land is to be sold such that the dealer makes an overall profit of 10%?

Video Solution

Solution

Let length = $$l$$ ft and breadth = $$b$$ ft

ABCD is original plot. First square AEFD is sold of side = DF = $$b$$, => FC = $$(l - b)$$

After that square EBHG is sold of side GH = $$(l - b)$$

=> HC = $$b - (l - b) = (2b - l)$$

Perimeter of remaining land GHCF = $$2 \times [(l - b) + (2b - l)]$$

= $$2b$$

Perimeter of original land ABCD = $$2 (l + b)$$

Acc to ques

=> $$\frac{2b}{2 (l + b)} = \frac{3}{8}$$

=> $$8b = 3l + 3b$$

=> $$5b = 3l$$

Only one combination is possible, i.e. $$l = 5$$ and $$b = 3$$ ($$\because$$ It is given that : $$l < 2b$$)

=> Area of land = $$5 \times 3 = 15$$ sq. ft

=> Cost of land = $$15 \times 1000 = Rs. 15,000$$

For overall profit to be 10 %, S.P. = $$15,000 \times \frac{110}{100}$$

= $$Rs. 16,500$$

Side of AEFD = $$b = 3$$

=> S.P. of AEFD = $$3^2 \times 1200 = Rs. 10,800$$

Side of EBHG = $$(l - b) = 5 - 3 = 2$$

S.P. of EBHG = $$2^2 \times 1150 = Rs. 4,600$$

Length of remaining rectangular land GHCF = $$(l - b) = 5 - 3 = 2$$

Breadth = $$(2b - l) = 6 - 5 = 1$$

Let selling price per sq. ft of GHCF = $$Rs. x$$

S.P. of GHCF = $$2 \times 1 \times = Rs. 2x$$

$$\therefore$$ Total S.P.

=> $$10,800 + 4,600 + 2x = 16,500$$

=> $$2x = 16,500 - 15,400$$

=> $$x = \frac{1100}{2} = Rs. 550$$

CAT Content

XAT 2012 Question Paper

If 5 of the dot balls had been hit for 4s, and if two of the shots for which the batsman scored 3 runs each had fetched him one run instead, what would have been the central angle of the sector corresponding to the percentage of runs scored in 4s? {Use data from the previous question}

Carpenter Rajesh has a circular piece of plywood of diameter 30 feet. He has cut out two disks of diameter 20 feet and 10 feet. What is the diameter of the largest disk that can be cut out from the remaining portion of the plywood piece?

Related Formulas With Tests

XAT Quant Questions | XAT Quantitative Ability

XAT DILR Questions | LRDI Questions For XAT

XAT Verbal Ability Questions | VARC Questions For XAT

Also Read

CAT Content