Test: CAT Quant 2019 ( Shift -1) - CAT MCQ

Minimum time taken by 1st car = 6 hours,

For maximum difference in time taken by both of them, car 1 has to start at 10:00 AM and car 2 has to start at 11:00 AM.

Hence, car 2 will take 5 hours.

Hence a= d/6 and b = d/5

Hence the speed of car 2 will exceed the speed of car 1 by (d/5 – d/6)/d/6X 100

= (d/30 / d/6) x 100 = 20

Now, 1/ √a₁ +√a₂ = √a₂ +√a₁ / (√a₂+√a₁) (√a₂+√a₁)

(Multiplying numerator and denominator by (√a₂ - √a₁

= √a₂ + √a₃ / (a₂ – a₁)

= √a₂ + √a₁ /d

(where d is the common difference)

Similarly, 1/ √a₂ + √a₃ = (√a₃ +√a₂)/d and so on.

Then the expression 1/√a₁ + √a₂ +1/√a₂+√a₃ …….. + 1/√a_n + √a_n+1 can be written as

1/d (√a2 - √a1 + √a3 - √a3 + …………. √an+1 - √an)

=n/nd ((√an+1 - (√a1)

= n(√an+1 - √a1)/ a_n+1 - a₁ [(an+1 - an = nd)]

= n/(√a₁+ √an+1)

Since AB is a diameter, AQB and APB’ will right angles.

In right triangle APB,

Now, 2AQ = AP

=>AQ= 8/2=4

In right triangle AQB,

Taking log in base 10 on both sides,

x(log₁₀ 555-2) = y(log₁₀ 555-3) = 3

Then, x(log₁₀555-2) = 3…..(1)

y(log₁₀555-3) = 3 ……(2)

From (1) and (2)

=>log₁₀555= 3/x + 2 = 3/y +3

=> 1/x – 1/y = 1/3

And the income of Kamala will be 120a*100/80=150a

If Kamala's income goes down by, 4%, then new income of Kamala = 150a-150a(4/100) = 150a-6a=144a

If Bimla's income goes up by 10 percent, her new income will be 100a+100a(10/100)=110a

=> Hence the Kamala income will exceed Bimla income by (144a-110a)*100/110a=31

Distance covered by B in 1 revolution = 2π*40 = 80π

Now, (5000+n)60π = 80πn

=> 15000= 4n-3n =>n=15000

Then distance travelled by B = 15000*80π cm = 12πkm

Hence, the speed = (12π X 60)/45= 16π

=>|(x-3)(x+2)|=x+2

For x<-2, (3-x)(-x-2)=x+2

=> x-3=1 =>x = 4 (Rejected as x<-2)

For -2 ≤ x<3, (3-x)(x+2)=x+2

=>x=2,-2

For x≥ 3, (x-3)(x+2)=x+2 =>x=4

Hence the product =4*-2*2=-16

Assuming the length of race course = x and the speed of three horses be a,b and c respectively.

Hence, x/a = (x-11)/b…. (1)

and x/a = (x – 90)/c …. (2)

Also, x/b = x – 80/c …. (3)

From 1 and 2, we get, (x-11)/b = (x-90)/c …. (4)

Dividing (3) by (4), we get, x – 11/x = x – 90/x-80

⇒ (x-11 )(x-80) = x(x-90)

⇒ 91x-90x=880
⇒ x=880

The population of town at the beginning of 2nd year = 3+2p

The population of town at the beginning of 3rd year = 2(3+2p)+3 = 2*2p+2*3+3 =4p+3(1 +2)

The population of town at the beginning of 4th year = 2(2*2p+2*3+3)+3 = 8p+3(1+2+4)

Similarly population at the beginning of the nth year = 2^n-1p+3(2^n-1- 1) = 2^n-1(p + 3)-3

The population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be (2^2034-2019) (1000 + 3)-3 = 2¹⁵ (1003)-3

f(x + y)=f(x)f(y)

Hence, f(2) = f(1+1)=f(1)*f(1)=2*2=4

f(3)=f(2+1)=f(2)*f(1)=4*2=8

f(4)=f(3+1)=f(3)*f(1)=8*2=16

……… =>f(x)=2^x

Now,f(a + 1)+f(a + 2) + ... + f(a + n) = 16(2ⁿ-1)

On putting n=1 in the equation we get, f(a+1)=16

⇒ f(a)*f(1)=16 (It is given that f (x + y) = f (x) f (y))

⇒ 2^a*2=16

⇒ a=3

Given, Bina's interest income exceeds Amala by 20x-18x=2x=250 => x=125

Now, total interest income = 18x+20x+20x=58x = 58*125 = 7250

Assuming m is even, then 8f(m+1)-f(m)=2

m+1 will be odd

So, 8(m+1+3)-m(m+1)=2

=> 8m+32-m²– m=2

=> m² − 7m − 30 = 0

=> m=10,-3

Rejecting the negative value, we get m=10

Assuming m is odd, m+1 will be even.

Then, 8(m+1) (m+2) – m – 3=2

=> 8(m²+3m+2) – m – 3=2

=> 8m² + 23m +11=0

Discriminant for the equation = 23²–4.8.11 = 177

Hence, the value of m will not be integral. Hence this case will be rejected.

We have, a³ – b³ / (a - b)³ = 157/3

=> 3 (a³ − b³) = 157 (a³ − b³ + 3ab (b − a))

=> 154 (a³ — b³) + 3 * 157 * ab(b — a) = 0

=> 154 (a³ - b³) + 3 * 616 * 157 (b - a) = 0 (ab=616)

=>a³ - b³ + (3 x 4 x 157 (b - a)) (154*4=616)

=> (a - b) (a² + b² + ab) = 3 x 4 x 157 (a - b)

=>a² + b² + ab = 3x4x157

Adding ab=616 on both sides, we get

a² + b² + ab + ab = 3 x 4 x 157 + 616

=> (a + b)² = 3 x 4 x 157 + 616 = 2500

=>a+b=50

Since Amal travelled 1/3rd of his total journey time in different speeds

d = t/3 x 10 + t/3 x 20 + t/3X 30 = 20t

Total time taken by Bimal = d₁/s₁ + d₂/s₂ + d₃/s₃

= 20t/3 x 1/10 + 20t/3 x 1/20 + 20t/3 x 1/30 = 20t(6+3+2)/3x30 = 11/9t

Hence, the ratio of time taken by Bimal to time taken by Amal =11t/(9/t)= 9

Therefore, Bimal will exceed Amal's time by (11t/9 – t)/tx 100 = 22.22

After increasing her score by 50%, she will get 40a(1 +50/100)-60a

Passing score = 60a+35

Post review score after 20% increase = 60a*1.2=72a

=>Hence, 60a+35+7=72a

=>12a=42

=>a=3.5

=> maximum marks = 350 and passing marks = 210+35=245

=> Passing percentage = 245*100/350 = 70

Hence, the interest = 200x*4/100 = 8x and 100x*3/100=3x

Interest from the fixed deposit = (1500000-300x)*6/100 = 90000-18x

Hence the total interest = 90000-18x+8x+3x=90000-7x =76000

=> 7x=14000

=>x=2000

Hence, the fixed deposit investment = 1500000 – 300*2000 = 900000 = 9 lakhs

The number of members who can play exactly 1 game = II

The number of members who can play exactly 1 game = III

1+2II+3III = 144+123+132=399....(1)

I+II+III=256……(2)

=>11+2111=143…..(3)

Also, II+3III=58+25+63=146…..(4)

=> III = 3 (From 3 and 4)

=> II = 137

=> I = 116

The members who play only tennis = 123 – 58 – 25+3 = 43

= 3(2¹² - 2 - 2¹¹ + 2) = 3(2¹¹)(2 - 1)= 3*2¹¹ = 6144

Positive Mark: 3

Negative Mark: 0

Hence, 12(a+b)=1……(1)

Also, 9(a/2+3b)=1

Using both equations, we get, 12(a+b)= 9(a/2+3b)

=> 4a+4b= 3a/2 +9b

5a/2 =5b

=> a=2b

Substituting the value of b in equation (1),

12(3a/2) = 1

=>a=1/18

Hence, the number of days required = 1/(1/18)=18

Hence, the number of girls = 60x and the number of boys = 40x

We have, 60x-40x=30 =>x=1.5

The number of girls = 60*1.5=90

Number of girls that pass = 68x – 30=68*1.5-30 = 102 – 30=72

The number of girls who do not pass = 90 – 72=18

Hence the percentage of girls who do not pass = 1800/90-20

B In figure AE*BE=CE*DE (The intersecting chords theorem)

=> 7*15=x(20.5-x)(Assuming AE=x)

=> 210=x(41 -2x)

=> 2x²-41x+210=0

=> x=10 or x=10.5

=> AE=10 or AE=10.5

Hence BE = 20.5-10=10.5 or BE = 20.5-10.5=10

Required difference= 10.5-10=0.5

So, on selling a pen at 5% loss and a book at 15% gain, net gain = -5p+15b = 7 ....1

On selling the pen at 5% gain and the book,at 10% gain, net gain = 5p+10b = 13…2

Adding 1 and 2 we get, -25b=20

Hence 100b= 20*4=80,

C is the answer.

The weight/volume(g/L) for liquid 2 = 800

The weight/volume(g/L) of the mixture = 480/(1/2) = 960

Using alligation the ratio of liquid 1 and liquid 2 in the mixture = (960-800)/(1000-960) = 160/40 = 4:1

Hence the percentage of liquid 1 in the mixture = 4*100/(4+1 )=80

Sum of the scores of 21 students other than Ramesh = 21 a

Hence the average of 22 students = a+1

Sum of the scores of all 22 students = 22(a+1)

The score of Ramesh = Sum of scores of all 22 students - Sum of the scores of 21 students other than Ramesh = 22(a+1)-21 a=a+22 = 82.5 (Given)

=> a = 60.5

Hence, sum of the scores of all 22 students = 22(a+1) = 22*61.5 = 1353

Now the sum of the scores of students other than Gautam = 21*62 = 1302

Hence the score of Gautam = 1353 – 1302=51

We have, (√2 )¹⁹3⁴4²9^m8ⁿ = 3ⁿ16^m(√64 )

Converting both sides in powers of 2 and 3, we get

2^19/23⁴2³3^2m2³ⁿ = 3ⁿ2^4m2^6/4Comparing the power of 2 we get, 19/2 + 4 + 3n = 4m + 6/4

=> 4m=3n+12…. (1)

Comparing the power of 3 we get, 4 + 2m = n

Substituting the value of n in (1), we get

4m=3(4+2m)+12

=> m=-12

Since, three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job, hence the efficiency will be double.

=> 3a+8b = 2(3b+8a)

=>13a=2b

Hence work done by 13 men in a day = work done by 2 machines in a day.

=> If two machines can finish the job in 13 days, then same work will be done 13 men in 13 days.

Hence the required number of men = 13

The given figure can be divided into 9 regions or equilateral triangles of equal areas as shown below,

Now the hexagon consists of 6 regions and the triangle consists of 9 regions.

Hence the ratio of areas = 6/9 =2:3

We have, log₅ (x + y) + log₅ (x - y) = 3

⇒ x² - y² = 125…..1)

log₂y - log₂X = 1 - log₂3

⇒ y/x = 2/3

⇒ 2x = 3y ⇒ x = 3y/2

On substituting the value of x in 1, we get

5x²/4 = 125

⇒ y = 10, x = 15

Hence xy = 150

Sum of the area of region I and II is the required area.

Now, required area = 4x 1/2 x 1 x 1=2