# CAT 2019 Quant Question Paper with Solution (Slot 1)

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## CAT 2019 Quant Solution

Q.1) Let x and y be positive real numbers such that
log5 (x + y) + log5 ( y) = 3, and log2 y − log2 x = 1 − log2 3. Then xy equals

a)250

b)100

c)150

d)25

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]c)150[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]

Solution:

Given

log5 (x + y) + log5 ( y) = 3  and

log2 y − log2 x = 1 − log2 3.

log5 (x + y) + log5 ( y) = log5 (x^2-y^2) = 3

=>x^2-y^2=5^3…..eq1

log2 y − log2 x = 1 − log2 3

log2 (y/x) = log2 2 − log2 3

log2 (y/x) = log2 2/3

y/x=2/3

x=3y/2…..putting this in eq1

9y^2/4-y^2=125

5y^2/4=125

y^2=100

y=10

x=15

xy=150

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Q.2) In a class, 60% of the students are girls and the rest are boys. There are 30 more girls than boys. If 68% of the students, including 30 boys, pass an examination, the percentage of the girls who do not pass is

a)20%

b)30%

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)20%[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]

Solution:

G girls

B boys

Let G=60x

B=40x

40x+30=60x

X=1.5

So, G=90

B=60

Total passed=150*68/100=102

Among 102 passed students, 30 are passed boys

So girls passed=102-30=72

So girls failed=total girls-girls passed=90-72=18

So % of girls failed=(18/90)*100=20%

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Q.3) A person invested a total amount of Rs 15 lakh. A part of it was invested in a fixed deposit earning 6% annual interest, and the remaining amount was invested in two other deposits in the ratio 2 : 1, earning annual interest at the rates of 4% and 3%, respectively. If the total annual interest income is Rs 76000 then the amount (in Rs lakh) invested in the fixed deposit was

a)9 lakh

b)8 lakh

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)9 lakh[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]

Solution:

Total =15 lakh

Let the amount invested in fixed deposit be = x at 6% SI

Remaining amount= 15-x ….which was invested in 2:1 at rates 4% and 3% per annum.

So amount invested at 4% pa=2/3(15-x)

Amount invested at 3% pa=1/3(15-x)

Total interest after 1 year=76000

So, (x*6*1)/100 + [2/3(15-x)*4*1]/100 + [1/3(15-x)*3*1]/100 = 76000

X=9 lakh

So 9 lakh will be the answer.

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Q.4) 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. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?

a)13

b)23

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)13 lakh[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]

Solution:

X denotes men

M denotes machines.

Work be W

(3X+8M)*t/2=W

(8X+3M)*t=W

Dividing both equations

(3X+8M)/(8X+3M)=2

3X+8M=16X+6M

13X=2M

Also given

2M*13=W

So, W=26M

Let K men finish Work W in 13 days

So, W/(K*X)=13…

26M/(K*2M/13)=13

K=13 [/bg_collapse]

Q.5) The income of Amala is 20% more than that of Bimala and 20% less than that of Kamala. If Kamala’s income goes down by 4% and Bimala’s goes up by 10%, then the percentage by which Kamala’s income would exceed Bimala’s is nearest to

a)31%

b)24%

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)31%[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]

Solution:

A=1.2B

A=0.8K

If K now becomes 0.96K and B now becomes 1.1B

Since B and K have a relation with A so, let us equate them

1.2B=0.8K

K=1.5B

So (New K -New B)/New B= ?

(0.96K-1.1B)/1.1B=[(1.44B-1.1B)/1.1B] * 100 = 0.34B/1.1B * 100=30.9 %

Which is close to 31% [/bg_collapse]

Q.6) Corners are cut off from an equilateral triangle T to produce a regular hexagon H. Then, the ratio of the area of H to the area of T is

a)2/3

b)2/4

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)2/3[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]

Solution:

We can see that the equilateral triangle is made up of 9 equal triangles

Hexagon is made up of 6 equal triangles of same size

So ratio of areas = 6/9=2/3

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Q.7) At their usual efficiency levels, A and B together finish a task in 12 days. If A had worked half as efficiently as she usually does, and B had worked thrice as efficiently as he usually does, the task would have been completed in 9 days. How many days would A take to finish the task if she works alone at her usual efficiency?

a)8

b)6

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)8[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]

Solution:

Let W be the work and a,b be the efficiencies of a and b

12(a+b)=W…..eq1

If A works at a/2 and b work at 3b per day.

Then 9(a/2+3b)=W…..eq2

Equating  equations 1 and 2

12a+12b=9a/2+27b

15a/2=15b

a=2b

so we need W/a

putting a=2b in eq1

W/a=8

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Q.8)

a)

b)

c)

d)

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]

Solution: