CAT Quant Mock Test - 3 - CAT MCQ

If 4 < x<4.5, g(x) = x- 4 + 4.5 -x + 4.8 - x = 5.3 - x

If 4.5

g(x) decreases with x for x < 4.5 and increases with x for 4.5 < x

∴g(x) has minimum value at x = 4.5

Since the sum is zero the number of terms equal to 1 must be same as the number of terms equal to -1. This is possible only when ‘n’ is even.

Hence, n must be an even number.

Considering 90 as the capacity of the tank {LCM of (45, 30, 15, 10, 9) we can find therate of flow for the 5 pipes as 90 / 45,90 / 30, 90 / 15, 90 / 10 and 90 / 9 respectively. The time and the rates of the 5 pipes are tabulated below.

If the tank gets filled in 12  6 / 7 min, effective rate of filing

= 90 / (90 / 7)= 7 units/min

If the tank gets emptied in 22 x 1 / 2min, effective rate of emptying = 90 / (45 / 2)= 4 units/min

7 units/min can be obtained by making P₄ a filling pipe and P₁ emptying pipe or Pe filling and P₂ emptying. If P₄ is filling pipe and P₁ emptying pipe, 4 units/min can be obtained by making P₅ emptying pipe and P₃ filling pipe. In the latter case, P₃ emptying and P₁ filling pipe.

Case(i):7 = 10-3 = P₅-P₂

-4 = 2 - 6 = P₁ - P₃

Case(ii):7 = 9- 2 = P₄ – P₁

-4 = 6- 10 = P₃-P₅

∴P₁, P₅ or P₃, P₂ may have been converted to emptying pipes. Only P2, P3 is listed among the options.

Further, the sign ofagiven expression oftheform ax + by can be determined for some cases of (a, b), where a, b are obtained by multiplying the first and second inequalities by p and q and adding the resulting inequalities (in the given question a = 2p - q and b- 3p + q) provided p, q have the same sign.

For the three given expressions, p, qr are tabulated below.

We conclude that the sign of 3x + 7y and 4x + 11 y can be determined but not that of 5(x + y) or x + y.

As N = 15, 14 ≤ e ≤ 105

Median = T₄ + T₅ / 2= m

=> T₄ + T₅ = 2m → (1)

When T₉, a number greater than T₈, is included, the median is T₅.

∴T₅ = m + 4→ (2)

when T₈ is removed, the median is T₄.

∴T₄ = m / 3 → (3)

From (1), (2) and (3), we get

m + 4+ m / 2 = 2m=>4 = m / 2 => m = 8

When a number smaller than the least number is added, median = T₄= m / 2 = 4

Let the speeds of Ram Kumar and Kishore be x steps per second and y steps per second respectively. Both having started simultaneously, the total length covered by them to cross each other must be equal to the total length of the escalator.

∴(x + y)21 = (x + y / 3) 28 = Number of steps on the escalator

∴ 9(x + y) = 4(3x + y)

5y = 3x

Y / x = 3 / 5

Let x = 5k and y = 3k,

Total steps in the escalator = 21 (5k + 3k) = 168k

Both having started from the ground floor in the time Ram kumar takes 5k₁steps, Kishore will take 3k₁ steps. So they will be separated by 2k₁ steps.

Total steps taken by Kishore to reach the 1^stfloor = 5K₁ – 20 / 100 = 5k₁ = 4k₁

So in the time Kishore took kt steps, the escalator took the remaining (2k₁– k₁) steps.

Thus speed of Kishore = Speed of the escalator.

Speed of the escalator = 3k steps/seconds

Since for half the time Ram Kumar will be walking in opposite direction to that of the escalator, relative speed of Ram Kumar = 5k - 3k = 2k steps/seconds.

For the other half of the time, with the escalator being at rest Ram Kumar’s speed will be 5k steps/seconds.

Let Ram Kumar cover n steps at 2k steps/seconds speed and the remaining at 5k steps/seconds speed.

Since the time taken is same in both cases, n / 2k = 168k – n / 5k

5 n = 336k - 2n

7n = 336k or n = 48k

Time taken by Ram Kumar to reach the ground floor = 2 x n / 2k = 2 x 48k / 2k= 48 seconds

=> Each of the four elements of B must have at least one corresponding element of A.

Two elements of A are mapped to one element of B. The remaining elements are mapped one each to the remaining elements of B.

Two elements from A can be selected in ⁵C₂ ways and one element from B can be selected in ⁴C₁ ways. The remaining three elements can be mapped in 3! ways.

Total ways = ⁵C₂. ⁴C₁ . 3! = 240

Given,

x + y + z = 35 → (1)

10x + 8y + 15z = 422 → (2)

8x + 8y + 8z = 280→ (3)

Subtracting (3) from (2), we get

2x + 7z = 142

The possible values of x and corresponding values of z have been substituted in the equation

x z y

142 = 2(15)+ 7(16) 4

= 2(8)+ 7(18)9

= 2(1)+ 7(20) 14

As he bought as many type C pens as possible (i.e. 20), he bought 15 pens of type A and B.

(i)less than 70%

(ii)more than 70% but not more than 90% (Distinction)

(iii)More than 90% (Distinction + certificate of Merit)

Hence each student can be selected in 3 ways.

For n students = 3x3x ……..n times = 3ⁿ

As EF is parallel to GH, △AEF and △AGH are similar. As the area of each part is equal, let area of △AEF be x. Then area of △AGH = 4x

Sum with A = nx + x/3

Had the amount left with A been twice more than what it was, then he could have bought 2 (x) x one more gold coin i.e., had his amount left been 2(x) / 3 + x / 3 = x, he could have bought another coin.

Sum with B = 7 times the sum with A = 7(nx + x/3)

= 7nx + 2x + x/3

Now x/3 + 7500 = x [ ∵with an extra ₹7500, B would have been able to purchase another 2x gold coin] or, 2x / 3 = 7500 or, x = ₹11,250

Therefore the cost of each gold coin was ₹11,250.

The line EF is the fold, which is made such that the corner D meets the diagonally opposite comer B. As DF coincides with FB upon making the fold DF = FB. Similarly, DE = EB. Also, as DF is the part of the length of the rectangle that is being folded so that D coincides with the opposite vertex and BE is the part of the length, that is being folded so that B coincides with D, DF = BE (from symmetry), i.e., quadrilateral DFBE is a rhombus (I) Now assume, FC = x cm

In the right angled triangle △BFC, BF =√BC² + FC² = √(30)² + x² = DF(since, EDBFis a rhombus and BF = DF)

Hence√30² + x² = 40 - x

=> 900+x²= 1600 + x²– 80x

=> x = 700 / 80 = 35 / 4

Now, consider G on DC, such that EG ⊥DC. In △EGF, GF = 40 - 2x (as AE = FC = x), EG = 30 and EF is the length of the fold.

Alternative Solution

Consider the conclusion (I), i.e., that EDBF is a rhombus. Let the diagonals of the rhombus meet at O. In a rhombus, the diagonals bisect each other at right angles. Hence △EOB is similar to △DAB (both are right angled, with a common angle at B).

Part of the job doneon the second day =(2+ 3 +…+M)units.

Part of the job done on the mlh day = (m +…+ M) units.

And finally, part of the job done on the Mth day = M units

Job =1+2+ 3+ + M

+ 2 + 3 +….+ M

+ 3……….

+ M

= (1² + 2² + 3²…..M²) units

= (M)(M + 1)(2M + 1) / 6 units

=>(M)(M + 1)(2M + 1) / 6 = 204 (given)

=> (M) (M + 1) (2M + 1) = 1224 = (8) (9) (17)

comparing both sides, M = 8

If x = y and x² + y² = 1800 => x= 30 and y = 30 so maximum value of x + y is 60

Given N_i = N_i-i + N_{i+ i}

Hence the number with 1137^thstudent

= N₁₁₃₇ = N₁₁₃₈ + N₁₁₃₈

=> 16 = -57 + N₁₁₃₃

=> N₁₁₃₈ = 73

similarly N₁₁₃₈ =-73

and N₁₁₃₉ = 57

and N₁₁₃₄ = -16

and N₁₁₄₀ = -16

∴we see that the series of numbers from N₁₁₃₄ onwards is as follows

-16, -73, -57, 16, 73, 57, -16, -73, -57 and so on, with the set of six values repeating continuously. Note that the sum of these six values themselves is zero.

From the pattern, we can say that the number with the student of the form N = 6k + 1 will be -73.

(∴ N₁₁₃₅ = -73 and 1135 = 6k + 1)

∴we can ignore all students till the highest multiple of six under (or equal to) 2272 i.e, till 2268.

Now only four more students will remain, and they will have the numbers-73,-57,16 and 73.

Hence, the sum of the numbers with them will be -41.

a₃ = a₄ = bh,

a₅ =a₆ =hl.

A₁ a₂a₃ a₄a₅ a₆ =ℓ⁴ b⁴ h⁴ ={ℓbh)⁴

∴ volume =

p²> 4q. If q = 1, p can be 2 or 3 or 4.

If q = 2, p can be 3 or 4.

If q = 3, p can be only 4.

If q = 4, p can be only 4.

∴There are 7 possible equations.

Let the number of students in each section be 4k.

The sum of the heights of all the students of section A is 4k x 155 = 620 k

The sum of the heights of all the students in section A after 1/4^thof the students of section B moved to section A is 5k x 156 = 780k

The sum of the heights of all the students who moved to section A from section B.

780k-620k = 160k

The average height of the students who moved = 160

= 160 cm

As the average height of the students who moved out of section B is the same as the average height of section B, the average height of section B will not change after the movement of students.

Let the value of 1 + 4x² + 9X⁴ + 16x + ... ∞ be denoted by S.

S = 1+4x² + 9x⁴+16x⁶ + … (1)

Sx² = x² + 4x⁴ + 9x⁶ + … (2)

Subtracting (2) from (1)

S (1 - x²) = 1 + 3x² + 5x⁴ + 7x⁶…(3)

Multiplying (3) by x², we have

Sx² (1 - x²) = x² + 3X⁴ + 5x⁶ + … (4)

Subtracting (4) from (3),

S (1 - x² - x² + x⁴) = 1 + 2x² + 2X⁴ + 2X⁸ + ..

⇒ 2n(2n + 1) / 1 = n(n + 1)(2n + 1) / 6 ⇒ n + 1 = 6 ⇒ n = 5

∴S = 5(11) = 55 and the remainder when S is divided by 2n(i.e., 10) is 5.

The length of the pipe is irrelevant information for the given question.

= 16 / 9

∴V₁ = 16V / 9

∴M ^{1 / 4} N⁴ = 4 => MN¹⁶ = 256.

The given data can be written as:

(y – k) = 1 / 2 (y + k) --- (1)

(y + P) = 3 (y-p) --- (2)

From (1), y + k / y – k = 2 / 1;

=> y / k = 3 / 1 (by componendo and dividendo) ---(3)

From (2), y + p / y – p = 3 / 1 => y / p = 4 / 2 = 2 ---(4)

Dividing (4) by (3), y / p x k / y = 2 / 3 ; => k : p = 2 : 3