# Free DI and LR - 04 Practice Test - CAT

### Question 1

**In July 2005, total number of students learning music was what percent of students learning music under Mr. Andrews from April 2005 to march 2006.**

#### SOLUTION

Solution :B

7.68%

If the no of students in July 2005 is x, then (16x100) = (7.23100)×(17100);x is slightly > 7.23.

### Question 2

**What is the approximate number of students learning Accordion in July 2005.**

#### SOLUTION

Solution :A

Number of students learning Accordion in 2005 = 0.01x = 0.01 × 7.68100 × 5270= 405.

### Question 3

**Which two instruments had the same ratio of the number of students in July 2005 to number of students in April 2005 to march 2006?**

#### SOLUTION

Solution :B

Accordion, Harmonica

The percentage of number of enrolments for Accordion and Harmonica were same in July 2005 and April 2005-March 2006.

### Question 4

**A total of 155500 students were enrolled from April 2005 to July 2005. What percent of students enrolled in a year, from April 2005 to march 2006, actually enrolled in the first quarter of the year?**

#### SOLUTION

Solution :B

The number of enrolments in the first quarter (April 2005 to June 2005) = 155500 – (7.68% of 527000) = 115000. Thus the required percentage = (115527)×100 = 22%.

### Question 5

**For which instrument was the ratio of no of enrolments in July 2005 to enrolments from April 2005 to March 2006 the maximum?**

#### SOLUTION

Solution :D

Looking at these charts, we find that Piano had the maximum ratio of enrolments in July 2005 compared to from April 2005 to March 2006.

### Question 6

What could the maximum percentage growth in marks in a single section be?

#### SOLUTION

Solution :C

Initially Gautam has scored 50 marks out of 100. So he has a 50% score.

Now the Test is changed to 25 marks for each section.

Hence, if he was supposed to get 15 marks in reasoning, if he exceeds this by 10 marks he will now get 25 out of 25. I. E 100 % marks

Or a increase of 100 %.

### Question 7

What is the maximum possible percentage growth in Gautam’s marks in Mock CEE 2 over Mock CEE 1?

#### SOLUTION

Solution :C

For the percentage increase in marks to be maximum, the increase in marks should also be maximum. This will happen only if he exceeds his expectations in the sections where he was expecting the maximum improvement

He cannot improve his VA percentage 3 times as that will cross 100%. He will improve his QA percentage by 3 times what he expected = 28×3 = 84% and he gets 10 marks more than expected in Reasoning

New scores in each section

Reasoning=25; QA=23; VA=18 and RC= 14

Total score on 100= 80. Initial score on 100 = 50

percentage growth = 60%

### Question 8

If the overall percentage increase in marks in Mock CEE 2 is 52%, then how much does he score in RC in Mock CEE 2?

#### SOLUTION

Solution :B

Since the percentage growth is given as 52%> marks = 12.5×1.52×4 =76 marks in Mock CEE2

The case will not be the previous case where the maximum growth is expected. This time it will be lesser.

If the percentage growth in his score in Reasoning is thrice that expected and he scores 10 marks more in RC, then his total score can be 78. We are looking for 76 marks

This will happen if the percentage growth in his score in RC is thrice that expected and he scores 10 marks more than expected in Reasoning. He then scores 25+16+18+17=76 marks. Thus, his percentage marks in RC= 12×3 = 36%

### Question 9

If Gautam had scored 5 marks more in each section than he had aimed at, then what would be his percentage growth in marks?

#### SOLUTION

Solution :D

Gautam would have scores 15+16+18+14+20= 83 on 100..this is a growth of 66%.

### Question 10

What could the minimum percentage growth in total marks in the two MOCK CEEs be?

#### SOLUTION

Solution :D

Minimum % growth happens when he exceeds his expectations in the subjects where he scored the least marks. REASONING=10 marks increase= 25(15+10).

QA= 16. VA=18 and RC=36% increase=19 Total = 78.

Percentage increase =23.8% .

### Question 11

How old is Blue?

#### SOLUTION

Solution :D

Let y1,y2......be the indefinite years. Let "t" be the year blue was born, "j" green and m yellow. Let y be the current year.

y−8−m=12(y2−j)=1−y3−ty3−m=5(y+2−t)...............(1)

y+10−t=2(y1−j)y1−m=9(y1−t)............(2)

t+1−m=3+y4−ty4−j=3(y5−6−m)

y5−j=12(y6−t)y6−m=10+y7−my7−j=13(y8−t)y8−m=3(j−m)

t=y−3j=y−8m=y−15Therefore, green = 8 years

blue= 3 years

yellow = 15 years

### Question 12

How old will Green be 10 years from now?

#### SOLUTION

Solution :C

Let y1,y2......be the indefinite years. Let "t" be the year blue was born, "j" green and m yellow. Let y be the current year.

y−8−m=12(y2−j)=1−y3−ty3−m=5(y+2−t)...............(1)

y+10−t=2(y1−j)y1−m=9(y1−t)............(2)

t+1−m=3+y4−ty4−j=3(y5−6−m)

y5−j=12(y6−t)y6−m=10+y7−my7−j=13(y8−t)y8−m=3(j−m)

t=y−3j=y−8m=y−15Therefore, green = 8 years

blue= 3 years

yellow = 15 years

### Question 13

How old would have Yellow been 6 years ago?

#### SOLUTION

Solution :C

Let y1,y2......be the indefinite years. Let "t" be the year blue was born, "j" green and m yellow. Let y be the current year.

y−8−m=12(y2−j)=1−y3−ty3−m=5(y+2−t)...............(1)

y+10−t=2(y1−j)y1−m=9(y1−t)............(2)

t+1−m=3+y4−ty4−j=3(y5−6−m)

y5−j=12(y6−t)y6−m=10+y7−my7−j=13(y8−t)y8−m=3(j−m)

t=y−3j=y−8m=y−15Therefore, green = 8 years

blue= 3 years

yellow = 15 years

### Question 14

Who is to the north-east of the person who is to the left of Amita ?

#### SOLUTION

Solution :D

To the left of Amita is Binoy and Shama, north east of both is Payal.

### Question 15

If a boy walks from Binoy, meets Shama followed by Amita, Ariya and then Payal, how many metres has he walked if he has travelled the straight distance all through?

215 metres

185metres

155 metres

245 metres

#### SOLUTION

Solution :A

B-S(25)

S-AM(40)

AM-ARIYA(60)

ARIYA-P(90)

TOTAL=215

### Question 16

Who is to the south of the person who is to the north-east of Shama?

#### SOLUTION

Solution :D

North east of Shama- Payal

South of Payal- Amita and Ariya