XAT Quantitative Ability And DI MCQ Quiz - 1 - CAT MCQ

Let 48 = r * cos y and 20 = r sin y. 48 cos x - 20 sin x = r(cos y * cos x) - r{sin y * sin x) = r{cos (x + y)) 

r= 52
Hence, minimum value of the given expression is -52.
Hence, option 3.

Here, the total collection per product and the total sales per product, each occupy exactly one-half of the pie-chart; and the total collection/sales for all 5 products is given at the top of the corresponding half of the pie-chart.  When calculating the percentage of collection/sales for each product out of the total collection/sales for all products, the values given in the pie-chart should be divided by 180°, not 360°.
Consider the following example: 

For T.V.
Total sales = 82.8° = (82.8/180) x 100 = 46%. Total sales = 46% of 1 lakh units = 0.46 lakh units Total collection = 79.2° = (79.2/180) x 100 = 44%. Total collection = 44% of 10 crores = Rs. 440 lakhs Total cost incurred in production = 0.46 x 800 = Rs. 368 lakhs.
Total profit = 440 - 368 = Rs. 72 lakhs Thus, the profit per unit can be calculated for each product as shown in the table above.
Profit earned on Fridge per unit = 28/0.16 = Rs. 175

Profit earned on Fridge and D.V.D is Rs. 28 lakhs and Rs. 10 lakhs respectively.
Required percentage = [(28 - 10)/10] * 100 = 180% Hence, option 1.

Total cost incurred = 368 + 152+135 + 130 + 104 = Rs. 889 lakhs Total collection = Rs. 1000 lakhs Profit percentage = [(1000 - 889)/889] * 100. = 12.48% ⇒ 12.5%
Hence, option 1. 

Note that only the magnitude of profit is to be considered to calculate the income tax.
Income tax for T.V. = [|72/440| x 100] = 16.36% Similarly, the income tax can be calculated for each product.
For Fridge = 15.55% For Computer = 12.5% For D.V.D. = 7.14% For Micro-wave oven = 13.3% Thus, the company pays the lowest income tax for D.V.Ds. Hence, option 4.

Consider the table obtained in the solution to the first question.
Total profit earned by the company = 72 + 2 8 - 1 5 + 10 + 16 = Rs. 111 lakhs.
Profit earned on Microwave ovens = Rs. 16 lakhs.
Therefore, share of Micro-wave ovens in the total profit = (16/111) x 100 = 14.4% Hence, option 4.

The first term of the series has 1 term, the second term has 3 terms, the third term has 5 terms and so on.
Therefore, the nth term will be having (2n - 1) terms and the first term of the nth term will be 1 more than the last term of the (n - 1)th term of the series.
The series is moving progressively taking every natural number starting with 1, so if we calculate how many terms have been used in all till the (n - 1)th term, we can find out the first term of the nth term which will be 1 more than the sum of all the terms used till the (n - 1)th term.

Number of terms till the (n - 1 )th term = 1 + 3 + 5 + ... + (2n-3)
= (n-1)²
The first term of the nth term = (n - 1 )² + 1 And number of terms in the nth term = 2n-1 .The last term of the nth term = (n - 1 )² + 1 + [(2n - 1) - 1 ] = (n - 1 )² + ( 2n - 1 ) = n²

^{Hence, option 1.}

First term of the nth term = (n - 1)² + 1 = n² + 2 - 2n, And number of terms in the nth term = 2n-1 The numbers that form the /7th term will be 

Let S be the smallest factor of N except 1 and L be the largest factor of N except N.
For any natural number N, Smallest factor * largest factor = 2nd smallest factor * 2nd largest factor = 3rd smallest factor x 3rd largest factor = ... = N Thus from the data given in the question and from the above rule, we get, S x L = N, L = 21S. Therefore  S x 21S = N. Therefore 21 S² = N
Since we want the largest factor (after N) to be 21 times the smallest factor (after 1), we will have to ensure that S takes values less than or equal to the smallest prime factor of 21 i.e. less than or equal to 3.

S = 2, 3
Two such numbers N are possible. And the two values of N are 21 x 2² = 84 and 21 x 3² = 189 Hence, option 2.

The probability that the coin will show tail = 1 - p. Let us first find the probability of getting exactly one head in 5 tosses.
Let only the first toss show head.
Probability o f this event = p x (1 - p)⁴ But second, third, fourth or fifth tosses can also show heads.
Hence, probability that there is exactly one head in 5 tosses = 5 x p x (1 - p)⁴

Similarly, let’s find the probability of getting exactly two heads in 5 tosses.
Let only the first and second tosses show head.
Probability = p² * (1 - p)³ Now, there are ⁵C₂ different ways in which we get exactly 2 heads in 5 tosses. 

Hence, probability that there are exactly 2 heads in 5 tosses = 5C₂ x p² x (1 - p)³ Hence, 5 x p x (1 - p)4 = ⁵C₂ x p² x (1 -p) 3
1 - p = 2p. 

p = 1/3
Now, let’s find the probability that there are exactly 3 heads in 5 tosses.
Let only first three tosses show head.
Hence, probability = p³ * (1 - p)² But there are 5C₃ different ways in which we can get exactly 3 head in 5 tosses.
Hence, probability of getting exactly 3 heads in 5 tosses = 5C₃ x p³ x (1 - p)²

= 40/243

Hence, option 3.

a + b + c + d = 1 =>b + c + d = 1 - a.  b + c + d is positive as b, c, and d are positive. (1 - a) is positive.
Similarly, (1 - b), (1 - c) and (1 - d) are positive as a + b + c + d = 1 and a, b, c, d are all positive.

So (1 - a)(1 - b)(1 - c)(1 - d) will be always less than both options 2 and 3.

Let A picks an integer x such that, 

A has to win and in step 4, since the result for A goes way above 2000, the game should end in step 3. i.e.,27x + 1300 >2000 i.e., 27x > 700 Least possible value of x is 26.
Also, for x = 26, 27x + 1200 is less than 2000.
Thus, x = 26 and the sum of its digits = 8 Hence, option 3.

We know that every prime number p > 3 can be expressed as 6k  1, where 'k' is a natural number.

3k²  k is always even irrespective of whether ‘k' is even or odd. Thus we conclude that ‘m’ will always be even. From this we can further conclude that ‘m’ can be expressed as 24n and thus from the above conclusion and from (2) and (3), we can say that every p² (where prime number p > 3) can be written as (24n + 1) for some natural number n.

∴ g(n) is prime for infinite values of n.

The number of 0's at the end of m depends on the number of occurrences of pair of 2 and 5 ( 2 x 5 = 10).
Now, m = 2⁵⁰ x (1 x 2 x ... x 49 x 50) m = 2⁵⁰ x 50! .

We can see that we have enough 2’s already and the constraint is on the occurrences of 5.
We calculate the number of occurrences of 5 in 50! as follows: [50/5] + [50/25] + [50/125] = 10 + 2 + 0=12 ...(Here, [x] represents greatest integer less than or equal to x). therefore 12 occurrences of 5 in 50!
There are 12 zeroes at the end of m.
Hence, option 3.

The fact that the fly has flown equal distances North and South shows that its final position is the same as its initial position; the point where the trains meet is 80 km from A.
The train starting from A has a speed of 40 km/hr, so it takes 2 hours to reach this point.
The other train must also take 2 hours to reach this same point, which is (200 - 80 =) 120 km from B, so the speed of the second train is 120/2 = 60 km/hr.
Hence, option 4.

There are 52 different type of cards i.e. 52 combinations of suit and denominations. Since no duplicate card is allowed, let us first choose 26 distinct type of cards from these 52 type of cards.
This can be done in ⁵²C₂₆ ways.
Now, when these 26 type of cards are selected, they could originally belong to any of the 3 packs of cards i.e. they could come from any of the 3 packs. This means that for each of the above 26 distinct type of cards, we have 3 ways of choosing them i.e. 3²⁶ ways.
Total number of ways = ⁵²C₂₆ x 3²⁶ Hence, option 3.

Refer the diagram, 

The shaded region represents the maximum area that can be grazed by the cow.
Let the cow be tethered to the top right corner (point B) of the rectangular house ABCD.
From the figure, it can be seen that:

i. The cow can graze 3/4th of the area of the circle of radius 20 ft.

ii. Moving to the left, along the length of the house the rope can go up to 16 ft along the boundary of the house. The remaining 4 ft of the rope would enable the cow to graze a small quarter circle of radius 4 ft. iii. Moving down, along the breath of the house, the rope can go up to 10 ft along the boundary of the house. The remaining 10 ft of the rope would enable the cow to graze another quarter circle of radius 10 ft.
The maximum total area that the cow can graze

The number of students in different specialisations in the age group of 25 years to 27 years in September 2007 and March 2008 are as shown in the table.
The number of students in each area of specialisation in March 2008 = Number of students in that area in September 2007 + Number of students in that area, who attain the age of 25 by March 2008 - Number of students in that area, who cross the age of 27 by March 2008 From the table, at least 24 - 15 = 9 students in marketing, 42 - 20 = 22 students in Consulting, and 12 - 10 = 2 students in HR crossed the age of 27 years between September 2007 and March 2008.
Hence, option 2.

To find the exact number of students in the age group 24 years 6 months to 25 years in September 2007, we need to know the number of students who crossed the age of 27 years in that period.
Hence, option E.

If the number of students in the age group 26 years 6 months to 27 years in September 2007 was the minimum possible then a minimum number of students would have crossed the age of 27 years (which is 33 from the first question).
It can be seen that only option 3 is correct.
Hence, option 3.

As the number of students who attained the age of 25 years was the minimum possible, it must be 2 (20 - 18).
Hence, option 2.

The number of students in each area of specialisation in March 2008 = Number of students in that area in September 2007 + Number of students in that area, who attain the age of 25 by March 2008 - Number of students in that area, who cross the age of 27 by March 2008 Consider Consulting, 20 = 42 + x - y. y - x = 22 The minimum value of y here is 22. Considering Marketing, 15 = 24 + x - y. y - x = 9

y has to be greater than 22.
Minimum value of y = 23 Minimum value of x= 14 Hence, option 2.

Let India’s population in 2010 be 100.
Hence, population of age group 41 to 45 years = 73.75 - 60.55 = 13.2
Hence, population of age group 41 to 50 was at least 13.2.
Now, population of age group 46 to 55 years is 88.65 - 73.75 = 14.9
Hence, maximum and minimum number of people in age group 46 to 50 is 14.9 and 0 respectively. Hence, maximum and minimum population of age group 41 to 50 years is 28.1 and 13.2 respectively.
Now, as population of only age group 41 to 50 increases, hence percentage increase in India’s total population is directly proportional to population of this group.
Hence, the required ratio = 13.2 : 28.1 « 0.469 : 1 Hence, option 2.

Let India's population, in 2010, was 100.
Hence,
Population of age group 31 to 40 years = 60.55 - 48.85 = 11.7
Population of age group more than 55 years = 100 - 88.65 = 11.35
Population of age group 16 to 20 years = 33.35 - 19.75 = 13.6
Population of age group upto 15 years = 19.75 Population of age group 41 to 45 years = 73.75 - 60.55 = 13.2

Hence, age group below 15 accounts for the highest population among the given age groups. Hence, option 4.

From the table, Population in the age-group of 56 and above accounts for 11.35% of total population.
Let ‘x' be the total population (in million)of the state.
Then, 11.35% of x = 25, x = 220.26. Population in the age-group of 15 - 20 years accounts for 13.6% (i.e. 33.35% - 19.75%) of total population.
Population in the age-group of 40 - 45 years accounts for 13.2 % (i.e. 73.75% - 60.55%) of total population.

Required difference (in million) = (13.6% of x) - (13.2% of x) = 0.4% of 220.26 

= 0.88 Or = 8.8 lakhs Hence, option 4.

From the table, Population of age-group upto 30 years accounts for 48.85% of total population.
Population of age-group above 55 years accounts for 11.35% of total population Required ratio = 48.85 : 11.35 ⇒ 1 : 0.232
Hence, option 4.

From statement A alone: When x = 4 and y = 0 , 2^x + 3^y= 2⁴ + 3⁰ = 16 + 1 = 17 and x + y = 4 + 0 = 4
When x = 3 and y = 2, 2^x + 3^y = 2³ + 3² = 8 + 9 = 17 and x + y = 3 + 2 = 5
In both the possibilities, x + y is not same.
Statement A alone is not sufficient as it does give us an unique answer.

From statement B alone: We have only one possibility.
When x = 3 and y = 2, 2^{(x + 2)} - 3^{(y- 2)}= 2⁽³⁺²) - 3^{(2 - 2)} = 2⁵ - 3⁰ = 32 - 1 = 31
x + y = 3 + 2 = 5. Statement B alone is sufficient to answer this question. Hence, option 2.

Let probability that Amar wins the match be a and probability that Baron wins the match is b. a + b = 1 ... (i) Baron will get a chance to shoot only if Amar has a blank shot.
Now, probability of Amar firing a blank shot is 5/6.
After Amar’s blank shot, the cylinder of the pistol is shuffled.

Hence, probability that Baron shoots a lethal shot is same as that of Amar. b = (5/6) x a Substituting this value in equation (i), we get, a + (5/6)a = 1 a = 6/11 Hence, b = 5/11 Hence, option 4.