CAT Quant Mock Test - 2 - CAT MCQ

Let the average weights of students in groups P and Q be p and q respectively. Total weights of students in groups P and Q are 20p and 30q respectively. Let the weight of the student who shifted from group P to group Q be w. After the shifting of the student,

the average weights of groups P and Q become 20p – w / 19 and 30q + w / 31respectively.

Given that20p – w / 19< p. Hence w > p.

Also given that 30q + w / 31> q. Hence w > q.

Alternative Solution

If a higher-average group joins or a lower-average group leaves, the average increases.

If a lower-average group joins or a higher-average group leaves, the average decreases.

The student leaves P and the average decreases. Therefore the student’s weight is more than initial (as well as the final) average weight of P. He joins Q and the average increases. Therefore, the student’s weight is more than the initial (as well as final) average weight of Q.

x² - 12x + 36 + y² - 6y + 9 + 28 = 36 + 9

(x - 6)² + (y - 3)² = 17

The above curve is a circle with centre at (6, 3) and radius √17 units. The least distance between the point and circle

= ∣Distance between the point and centre - Radius of the circle∣

Distance between the point (4, 0) and the center (6, 3) is√(6 – 4)² + (3 – 0)²

= √2² + 3² = √13

Radius = √17 units

=> least distance =√17 - √13 units.

The average of the annual CTC’s of the students who are placed in M, F and O is 13.

=> The sum of the annual CTC’s of the students who are placed in M, F, O is 13(m + f + o) → (1)

Similarly, the sum of the annual CTC's of the students who are placed in F, O, I is 14(f + o + i) → (2)

The sum of the annual CTC’s of the students who are placed in M, O, I is 12(m + o + i) → (3)

The sum of the annual CTC’s of the students who are placed in M, F, I is 15(m + f + i) → (4}

We can observe that each of the four different sectors appears in exactly three of (1), (2), (3) and (4).

∴The average annual CTC of the students placed

Since, m + 2i + 3f will be less than 3 ( m + f + o + i), then average annual CTC of the students will be between 13 and 14.

∴Among the given options only II is possible.

Given that for a variable ‘x’, ‘f’ is such that

Hence the function should be f(x) = x² - 2

The above figure gives the side view of the container and the balls.

Length of the box = 5 x 2r = 10r

Radius of the box = r

Volume of the box = πr² x 10r = π10r³

Total volume of the five balls = 5 x 4/3 x π x r³ = 20r³π / 3

Unoccupied volume = 10πr³(1 – 2/3) = 10πr³(1 /3).

∴Required ratio = 10πr³: 20/3 πr³ =1:2

Let the sum of frequencies initially considered be k

=> The actual sum of frequencies = k + 3

Calculated mean = 24

=> calculated sum of all observations = 24k

Actual sum = 24k + 3x

Actual mean = 24k + 3x / k + 3 = 24 + 5 = 29

=> 24k + 3x = 29k + 87

=> 5k = 3x - 87

=> k = 3(x – 29) / 5

As both k and x are integers, x - 29 must be a multiple of 5 and k must be a multiple of 3.

=> k + 3 must be a multiple of 3.

Among the given options, only 42 is a multiple of 3.

2 logx + 4 logx - 3 logy + 6 logx - 6 logy + 8 logx - 9 logy + + 2n logx - 3(n - 1) logy

Alternative Solution

This question can also be answered by eliminating the answer choices. By substituting n = 1, choices (b), (c) get eliminated and for n = 2; choice (d) is eliminated.

∴Only choice (a) can be the answer.

Manufacturing Expenses = 1,50,000

Overheads = 1,00,000

Total Expenses = 2,50,000

Profit = 7,50,000

Profit/Unit = 10,000

Reduction in profit = 35,000 35,000

∴Average number of idle units = 35,000 / 10,000 =3.5

2(ℓ + 0.75b) = 0.9 x (2(ℓ + b))

=> ℓ + 0.75b = 0.9ℓ + 0.9b

=>0.1 ℓ = 0.15b

=>ℓ : b = 3 :2

Let ℓ = 30 and b = 20

=> Present ℓ, b = (30, 15), i.e., after decrease in breadth.

Now, the length should be increased by 5 in order to compensate (i.e., retain ℓ + b = 30 + 20 = 50) for the decrease in breadth.

Hence 5 / 30 x 100 = 16.67%

The largest circle that can be cut from the quadrant must touch the two radii and the arc of the quadrant. Let its radius be r.

AOCB is a square where O is the centre of the circle to be cut.

The radius of the quadrant BD = BO + OD

= r√2 + r = 6

=> r = 6 / √2 + 1 = 6(√2 – 1)

Alternately, the radius of the smaller circle must be less than half the radius of the larger circle. Hence r < 3.="" by="" simple="" calculation,="" only="" choice="" (c)="" />

Now, (√a +√b +√c =a+b+c+2√ab +2√bc +2√ac

13+√48 + √72+ √96 = 13+2√12 + 2√18 + 2√24

Therefore the square root is

√a +√b +√c =√4+√3+√6

i.e. 2 + √3(√2+ 1)

Alternative Solution

Choices (A) and (C) can be easily eliminated since none of the surds in the given expression has √5 as factor. Choice (B) is also eliminated since √2x√8 would make the rational part much more (2 + 8 + 3 + 4= 17) than 13 (in the given expression). Hence, by elimination, Choice (d).

From I alone, we have

4r + 2w< 2r+4w

=> r < w.

It is not sufficient.

From n alone we have

2w + 4j = 4r + 2j => w+j= 2r

It is not sufficient

By combining both I and II, we can say that w> r> j.

Let us assume Robert bought x articles of each kind.

Total Amount spent = 8x + 25x + 38x = 2769

=> 71x = 2769

=> x = 39

∴Robert bought 39 articles of each kind.

=>A total of 117 articles.

= (cos²x + sin²x)³ - 3 cos²x sin²x (cos²x + sin²x)

For all x, 0 ≤ sin² 2x ≤ 1.

If sin² 2x = 0, then E = 1 and if sin² 2x= I, then E = 1 / 4

∴The required range is [1 / 4, 1]

Alternative Solution

Since sin⁶x + cos⁶x is always positive, choices (A) and (D) can be eliminated. Further, since both sinx and cosx cannot be simultaneously zero, even choice (C) is eliminated. Hence, choice (b).

Quantity of water in the first container = 60 litre.

Let the quantity of mixture added from the second container be 10x. 50 + 7x : 60 + 3x = 6:5

=>x= 6.471

Hence, 10x= 64.71 litres of mixture must be added.

If the speed of x km/hr be maintained for 100 - x minutes, the maximum distance that 50 can be covered occurs when x = 100 -x, i.e., x - 50 and it is 50 / 60 (50) = 41 x 2 / 3km

Given A - S = 24

=> A = S + 24 — (1)

Also, x years ago (A- x) = 3(S - x)

=> (S + 24 - x) = 2(S - x)— (2)

Similarly, 5x years ago

(S + 24 - 5x) = 3(S - 5x) — (3)

Solving (2) and (3), x = 3, S = 27 and A = 51=> S + A = 78

Alternative solution

Since the difference A - S is even (i.e., 24) the sum also must be even. We could check for choice (B) and (D). Hence choices (1) and (3) are eliminated.

Further, for choice (B)

A + S = 78 and A - S = 24 => A = 51, S = 27

By some observation. (51 - 3) = 2(27 - 3) and (51 - 15)

= 3(27 - 15) i.e., x = 3

Let the numbers of two-wheelers and four-wheelers at the parking lot be T and F respectively

T + F = 50 and 2T + 4F = 160.

Solving these, T = 20 and F = 30.

The total distance travelled by train in 50 sec.

= 50 x 72 x 5 / 18 = 1000 m.

∴ The length of the platform = 1000 - 600 = 400 m

He sold it at 225 x 1.2 = 270

Profit he made = 270 - 225 = 45

If he sold it the next day, the selling price increased by 15% (the increase in C.P. would not affect the merchant since he has already purchased the pack before the price change)

Next dayS.P. = 270 x 1.15 = 310.5

Profit if he sold it the next day = 310.5 - 225 = 85.5

Additional profit = 85.5 - 45 = 40.5

∴ Time taken by P and R to complete the same work is (80)(48) / 80 + 48 = 30 days.

for n = 8 (8)(7)(6) / 6 = 56 < 59

So minimum possible n is 9.

= Area of AOAP + Area of rectangle PBCQ + Area of triangle QCD

= (1 / 2 x 2 x 20) + (1 x 30) + (1 / 2 x 1 x 30) = 20 + 30 + 15 = 65 m

∴ Distance travelled is 65 m

The given lines 3x - 2y + 6 = 0 and 3x + 5y - 15 = 0 intersect on the y-axis at (0, 3). The area of the triangle formed by the lines 3x - 2y + 6 = 0, 3x + 5y - 15 = 0 and x-axis will be 1 / 2 x AO x BC = 1 / 2 x 3 x 7 = 10.5 sq.units. A = 10.5

∴4A =42