# Clat Exam Paper with Ans Keys: Section III-Numeric Ability

## Diagnostic Mock Test on CLAT Pattern

**SECTION III: NUMERIC ABILITY**

91. A sphere of radius 4 cm is melted and recast into a cylindrical shape. If the radius of the cylinder is 1 cm, find the height of the cylinder?

(a) 64/3 cm

(b) 16 cm

(c) 256/3 cm

(d) 48 cm

92. Two people are running around a track which is in the form of a square. The side of the square is 10 m. If they start from a point, move in the same direction and their speeds are 5 m/sec and 25 m/sec, after how much time from the start will they meet for the first time?

(a) 5 sec

(b) 1/3 sec

(c) 0.5 sec

(d) 2 sec

93. 20 litres of a 4:1 milk and water solution is mixed with 50 litres of a 2:3 milk and water solution. Find the ratio of milk to water in the resultant mixture?

(a) 18:17

(b) 18:35

(c) 1:1

(d) None of these

94. When the cost price and the selling price are both increased by x%, the value of profit changes from Rs 30 to Rs 39. Find the value of x?

(a) 10%

(b) 20%

(c) 30%

(d) 40%

95. In a class of 150 students, each student had to opt for at least one of the three subjects of Hindi, English or Computers. The number of students who opt for Hindi, English and Computers are 60, 60 and 59 respectively. 15 students opted for both English and Hindi, 10 students opted for both Computers and Hindi while

12 students opted for English and Computers. Find the number of students who opt for all the three subjects?

(a) 6

(b) 8

(c) 10

(d) 12

96. How many 4 digit odd numbers can be formed using the digits 0 to 7 without repetition such that the digit at the thousand’s place is also an odd digit?

(a) 360 ways

(b) 432 ways

(c) 300 ways

(d) 480 ways

97. In how many different ways can the alphabets of the word DECIDE be arranged such that the vowels are always together?

(a) 72 ways

(b) 144 ways

(c) 48 ways

(d) 36 ways

98. 15 men can make 30 identical objects in 15 days, in how many days can 45 men make 180 such identical objects?

(a) 45 days (b) 30 days

(c) 15 days (d) 60 days

99. A alone can do a work in 9 days more than that taken by A and B together to do the work while B alone can do the same work in 25 days more than that taken by A and B to do the work together. In how many days can A and B do the work together?

(a) 15 days (b) 20 days

(c) 12 days (d) 16 days

100. The sum of a two digit number and the number formed by interchanging the digits of the number is a multiple of 44. How many such numbers are possible?

(a) 5 (b) 6 (c) 3 (d) 4

101. Buses to destination C arrive after every 6 minutes while buses going to destination E arrive after every 8 minutes. If buses to both the destinations last arrived together at 1:12 p.m., when will buses to both the destinations together arrive again?

(a) 1:36 p.m. (b) 2:00 p.m.

(c) 2:12 p.m. (d) None of these

102. Which among the following will result in the maximum discount for a customer? (a) 2 discounts of 8% and 12%

(b) 2 discounts of 10% each

(c) Single discount of 20%

(d) All of them are equal

103. If 50% of a number is equal to 9.09% of 770, find the remainder when the number is divided by 3?

(a) 0 (b) 1

(c) 2 (d) cannot be determined

104. A reduction of 15% in the price of bananas enables a person to buy 15 bananas more for Rs 340. Find the number of bananas that can be purchased for Rs 340 after the price is reduced?

(a) 85 (b) 100 (c) 75 (d) 115

105. In a 75 m race A can beat B by 15 m while in a 60 m race B can beat C by 10 m.

By how many metres can A beat C in a 300 m race?

(a) 150 m (b) 100 m

(c) 175 m (d) 90 m

106. A train 150 m long and running at 72 kmph passes a man moving in the opposite direction in 5 seconds. Find the difference in their speeds in kmph?

(a) 72 kmph (b) 48 kmph

(c) 36 kmph (d) 60 kmph

107. In a class girls constitute 40% of the class strength. In an exam 50% of the girls qualified for the next round. If 30% of the entire class qualified for the next round, find the % of boys who qualified for the next round?

(a) 33.33% (b) 50%

(c) 20% (d) 16.66%

108. The ratio of the present age of a father and daughter is 8:3 while the ratio of the present age of the daughter and mother is 5:11. If the father’s present age is 40 years, find the mother’s age 2 years ago?

(a) 31 years (b) 33 years

(c) 35 years (d) 34 years

109. 72 litres of a solution contains milk and water in the ratio 5:3. Water is added to make the ratio 3:5. Find the quantity of water added?

(a) 36 litres (b) 40 litres

(c) 44 litres (d) 48 litres

110. The average score of a group of 6 students is 30 marks. When the marks of 2 of the 6 students are not added, the average marks of the remaining 4 students increases by 7.5 marks. Find the average marks of the 2 students whose marks are not added?

(a) 37.5 (b) 15

(c) 22.5 (d) None of these

**ANSWERS AND EXPLANATIONS**

**1**. (b) **2**. (c)

**3**. (d) **4**. (c)

**5**. (c) **6**. (a)

**7**. (b) **8**. (b)

**9**. (c) **10**. (d)

**11**. (d) **12**. (b)

**13**. (a) **14**. (b)

**15**. (d) **16**. (c)

**17**. (a) **18**. (b)

**19**. (c) **20**. (b)

**21**. (a) **22**. (b)

**23**. (d) **24**. (d)

**25**. (c) **26**. (b)

**27**. (d) **28**. (a)

**29**. (d) **30**. (c)

**31**. (c) **32**. (b)

**33**. (b) **34**. (d)

**35**. (d) **36**. (a)

**37**. (b) **38**. (b)

**39**. (b) **40**. (c)

**41**. (c) **42**. (b)

**43**. (b) **44**. (d)

**45**. (a) **46**. (d)

**47**. (a) **48**. (c)

**49**. (d) **50**. (c)

**51**. (a) **52**. (a)

**53**. (c) **54**. (b)

**55**. (b) **56**. (c)

**57**. (a) **58**. (a)

**59**. (c) **60**. (a)

**61**. (a) **62**. (d)

**63**. (b) **64**. (c)

**65**. (c) **66**. (a)

**67**. (a) **68**. (c)

**69**. (b) **70**. (a)

**71**. (c) **72**. (d)

**73**. (d) **74**. (c)

**75**. (d) **76**. (b)

**77**. (d) **78**. (d)

**79**. (b) **80**. (c)

**81**. (a) **82**. (a)

**83**. (d) **84**. (b)

**85**. (c) **86**. (a)

**87**. (d) **88**. (b)

**89**. (c) **90**. (b)

**91**. (c) **92**. (d)

**93**. (a) **94**. (c)

**95**. (d) **96**. (a)

**97**. (d) **98**. (b)

**99**. (a) **100**. (d)

**101**. (a) **102**. (c)

**103**. (c) **104**. (b)

**105**. (b) **106**. (c)

**107**. (d) **108**. (a)

**109**. (d) **110**. (b)

**111**. (c) **112**. (c)

**113**. (c) **114**. (c)

**115**. (b) **116**. (c)

**117**. (a) **118**. (b)

**119**. (d) **120**. (d)

**121**. (c) **122**. (c)

**123**. (a) **124**. (b)

**125**. (a) **126**. (c)

**127**. (d) **128**. (a)

**129**. (b) **130**. (a)

**131**. (b) **132**. (a)

**133**. (b) **134**. (c)

**135**. (b) **136**. (c)

**137**. (c) **138**. (a)

**139**. (b) **140**. (d)

**141**. (b) **142**. (c)

**143**. (c) **144**. (a)

**145**. (c) **146**. (b)

**147**. (b) **148**. (c)

**149**. (c) **150**. (a)

**151**. (a) **152**. (c)

**153**. (b) **154**. (c)

**155**. (c) **156**. (b)

**157**. (a) **158**. (d)

**159**. (b) **160**. (d)

**161**. (c) **162**. (a)

**163**. (d) **164**. (a)

**165**. (c) **166**. (b)

**167**. (d) **168**. (a)

**169**. (b) **170**. (c)

**171**. (a) **172**. (c)

**173**. (d) **174**. (c)

**175**. (d) **176**. (d)

**177**. (a) **178**. (c)

**179**. (d) **180**. (a)

**181**. (b) **182**. (b)

**183**. (c) **184**. (d)

**185**. (d) **186**. (b)

**187**. (d) **188**. (d)

**189**. (a) **190**. (b)

**191**. (c) **192**. (a)

**193**. (a) **194**. (b)

**195**. (b) **196**. (c)

**197**. (d) **198**. (b)

**199**. (a) **200**. (b)

**ANSWERS AND EXPLANATIONS**

91. (c) In this problem, the volume of the sphere is equal to the volume of the cylinder.

Volume of the sphere = (4/3) Πr3

Volume of the cylinder = Πr2h Therefore, (4/3)Πr3 = Πr2h (4/3) x 4 x 4 x 4 = 1 x 1 x h Therefore, h = 256/3 cm.

92. (d) For them to meet again after the start, the faster person must have covered one complete round more than the slower one.

One complete round would be 4 x side of the square = 4 x 10 = 40 m

Their relative speed = 25 m/sec – 5 m/sec = 20 m/sec

Time taken for them to meet again after the start = 40/20 = 2 sec.

93. (a) 20 litres 50 litres

4:1 2:3

Fraction 4/5 2/5 x

Let x be the resultant fraction of milk in the mixture. Therefore,

20/50 = (2/5 – x) / (x-4/5)

2x-8/5 = 2-5x

7x = 2+8/5

7x = 18/5 x = 18/35

Therefore, ratio of milk to water

= 18:17.

94. (c) We know that when the CP and SP are both increased or both decreased by the same %, there is no change in the % profit or loss. However, the absolute value of profit or loss increases or decreases by the same % by which both the CP and the SP have been increased or decreased.

Since [(39–30)/30] x 100 = 30%

Therefore, the CP and the SP must have been both increased by 30%.

95. (b) We can use the formula here n(AUBUC) = n(A)+n(B)+n(C)– n(AÇB)–n(AÇC)–n(BÇC)+ n(AÇBÇC)

150 = 60+60+59–15–10–12+x

where x represents the number of students who study all the three subjects.

150 = 179–37+x

150 = 142+x x=8.

96. (a)................3 ways............. x...............x............. x...............4 ways

The unit’s place can be filled with either 1, 3, 5 or 7, *i.e. *4 ways.

After doing this, the thousand’s place can be filled in 3 ways (since we have

already used one digit out of the 4 odd digits available). We had a total of 8 digits (0 to 7). We have used 2 digits.

The hundred’s place can be filled in 6 ways and the ten’s place in 5 ways.

Total number of ways = 3 x 6 x 5 x 4 = 18 x 20 = 360 ways.

97. (d) Let the vowels EIE be a group.

So, we have a group and 3 other alphabets D, C and D. These 4 things can be arranged in 4! / 2! ways (since the Ds are identical)

Also, the vowels can be arranged within the group in 3! / 2! ways

(since the 2 Es are identical) Total number of ways

= 4! x 3! / 2! X 2!

= 24 x 6 / 2 x 2

= 36 ways.

98. (b) As per the question, 15 men x 15 days = 30 objects------------(1)

45 men x y days = 180 objects-----------(2) Dividing (2) by (1) we have

45 x y / 15 x 15 = 180/30

45 x y = 15 x 15 x 6 y = 30 days.

99. (a) Let *A *and *B *do the work in x days.

Therefore, *A *alone can do it in (x+9) days while *B *alone can do it in (x+25)

days.

Let the total work be (x+9) (x+25) Efficiency of *A *= x+25

Efficiency of *B *= x+9

As per the problem, (x+25+x+9) x = (x+25)(x+9)

2x2 + 34x = x2+34 x + 225 x2=225

x=15 days.

Alternate method: If *A *takes x days more than *A *and *B *together to do a job while *B *takes y days more than *A *and *B *together to do the job, then the number of days for *A *and *B *to do the job together = √xy.

In this case, required value = √9x15 = √225 = 15 days.

100. (d) Let the 2 digit number be 10x+y and so the number obtained by interchanging the digits of the number will be 10y+x.

As per the problem, 10x+y+10y+x = 44M where M is any integer.

*i.e. *the value of 44M can be 44 or 88 or 132 or 176.

The left hand side = 11(x+y) *i.e. *x+y can be either be 4 or 8 or 12 or 16.

4 such values are possible.

101. (a) Buses to both the destinations will together arrive after a time which is equal to the LCM of 6 minutes and 8 minutes, *i.e. *24 minutes.

If both the buses arrived together at 1:12 p.m., they will again arrive after 24

minutes, *i.e. *at 1:36 p.m.

102. (c)

103. (c) 9.09% is the equivalent of 1/11.

Therefore, 9.09% of 770 = 1/11 of 770 = 70. This represents 50% of the number N. Therefore, the number N will be 140.

When 140 is divided by 3, the remainder obtained is 2.

104. (b) Let the initial price be 100 and so the reduced price will be 85.

If price changes from 100 to 85 and expenditure is constant, quantity purchased will change from 85 to 100, *i.e. *an increase of 15 which is the data given in the problem.

Therefore, the number of bananas that can be purchased after the price is

reduced = 100.

105. (b) *A *75 m *B *60 m *B *60 m *C *50 m

Since *B *is equal, we can compare *A *and *C*

In a 75 m race, *A *can beat *C *by 25 m and therefore in a 300 m race *A *can beat *C *by 100 m.

106. (c) Time taken for the train to cross the man = 150/5 = 30 m/sec Speed of the train = 72 kmph = 72 x 5/18 = 20 m/sec Therefore, speed of the man = 30–20 = 10 m/sec

Speed in kmph = 10 m/sec x 18/5 = 36 kmph.

So, the difference in speeds in kmph will be (72–36) = 36 kmph.

107. (d) Let the strength of the class be 100.

Number of girls = 40

Number of boys = 60

Total number of students who qualified = 30% of 100 = 30. Number of girls who qualified = 50% of 40 = 20.

Number of boys who qualified = 10.

% of boys who qualified = (10/60) x 100 = 100/6 = 16.66%.

108. (a) The ratio of the age of the father and daughter = 8:3.

If the father’s present age is 40, the daughter’s present age = 15.

Ratio of the age of the daughter and mother = 5:11

If the daughter’s present age is 15, the mother’s present age is 33. Mother’s age 2 years ago = 33–2 = 31 years.

109. (d) 72 litres of the solution contains milk and water in the ratio 5:3.

Quantity of milk = 5/8 x 72 = 45 litres. Quantity of water = 3/8 x 72 = 27 litres. Let x litres of water be added. Therefore,

45 / (27 + x) = 3/5

225 = 81+3x

3x = 144

x= 48 litres.

110. (b) Average score of the group of 6 students is 30 which means the total marks is

180.

Let the marks of the 2 students be y. Therefore, (180–y) / 4 = 37.5

180–y = 150

y = 180–150 = 30 marks.

Average marks of the 2 students = 30/2 = 15 marks.

**ANALYSE YOUR PERFORMANCE**

Mathematics

A medium section with questions from all aspects of maths, the questions were of fair difficulty with some blazers which would make the student think, nothing too tough though. A student with consistent maths practice and who has attended classes regularly would not have taken too long in this section. A score of 15+ would be considered a good score. Between 10–15 would be average and below that would require work on the part of the student.