# Free Quant Practice Test - CAT

### Question 1

How many pairs of co-prime natural numbers a and b exist such that a + b = 510?

60

120

64

128

#### SOLUTION

Solution :C

510=2×3×5×17.

Use the concept of Euler's theorem:

510(1−12)(1−13)(1−15)(1−117)=128

Number of pairs=1282=64.

### Question 2

There are some boxes labelled 1, 2, 3, 4... and so on up to 2n, where n>6 such that, for k = 1, 2, 3, 4, .... 2n, there are exactly k boxes labelled k. The total number of boxes is N. Now, consider the following two cases for the number of footballs (fi) in the box labelled i:

Case 1: fi=i+1, if i is odd; = i + 2 if i is even.

Case 2: fi=i+2, if i is odd; = i + 1 if i is even.

The difference between the total number of footballs in the N boxes in Case 1 and Case 2 is:

√N

√8N+1−14

N(N+1)6

√8N+1−12

#### SOLUTION

Solution :B

Method 1- Conventional Approach

Clearly, in both cases, fi=(i+1) or (i+1) + 1.

Therefore in both the cases, (i+1) is common.

As we need to find the difference in number of footballs, we can ignore (i+1).

Thus, effectively in each box, there will be 0 or 1 football.

We get:

Case 1: fi=0 if i is odd, = 1 if i is even.

Case 2: fi=1 if i is odd, = 0 if i is even.

Upon observation, we can conclude that the required difference is nothing but the difference in the number of boxes with odd numbered labels and the number of boxes with even numbered labels.

Also, number of boxes with label (x) is 1 more than number of boxes with label (x-1).

From 2n labels, we get n pairs of labels with consecutive numbers.

Therefore, required difference is n×1=n.

Or Case 1 - Case 2 = [{2+4+6+....+(2n)}-{1+3+5+....+(2n-1)}]

=2n(n+1)2−n2=n

Total number of boxes = 1+2+3+....+2n= n(2n+1) = N

Therefore, 2n2+n−N=0

n=−1±√1+8N4

Since n cannot be negative, n=(√8N+1)−14.

Therefore, answer option (b) is the correct answer choice.

Method 2- Assumption

This is a question based on Variables to Variables.

Assume n=1, hence the number of boxes labelled 1, will be 1 and the number of boxes labelled 2 will be 2. Total number of boxes (N) = 3.

Case 1

When i=1, f1=2 and when i=2, f2=4

Case 2

When i=1,f1=3 and when i=2, f2=3

Difference in total number of footballs in both cases = (2+4+4) - (3+3+3) = 1

Now, on substitution of N=3 in options, eliminate those options where you are not getting 1.

Only option (b) satisfies the required condition.

### Question 3

Find the minimum value of the function f(x) = |x-1| + |x-2| + |x-3| + ... + |x-20|

190

172

96

100

#### SOLUTION

Solution :D

The minimum value of such an expression occurs when x takes the middle value in the given range i.e. when x is between 10 and 11. And the value of the function f(x) at that value becomes:

f(x=10) = 9+8+7+6+5+4+3+2+1+0+1+2+3+4+5+6+7+8+9+10 = 100

Also,f(x=11) = 10+9+8+7+6+5+4+3+2+1+0+1+2+3+4+5+6+7+8+9=100Therefore, answer option (d) stands out as the correct answer choice.

### Question 4

Given that: Side of square ABCD = 4cm, E, F, G and H are midpoints of the sides AD, AB, BC and CD respectively. Also, the line passing through ON would bisect the line XF. Similar is the case for all other pairs: PO, RS and TM. Lastly, the lengths TM = NO = PQ = RS = 14 side of square (Note: The figure given below is a symmetric figure). X is the centre point of the square. The area of the shaded region is

#### SOLUTION

Solution :Using graphical division, we can divide the figure to look like the following:

Looking at the figure, we can easily make out that the shaded region is made out of 16 triangles,

Each being 1/64th the size of the larger square. (Since, area of each triangle =14 * 116 is area of square)

Therefore, area of shaded region = (14×116×16×16)cm2

Thus, required area of shaded region = 4cm2.

### Question 5

The king took a glass filled with red wine and drank 15 of its contents. When the king looked away, the court jester refilled the glass by adding water to the remaining red wine and then stirred it. The king drank 14 of this liquid mixture. When the king looked away again, the court jester refilled the cup with more water and stirred it. The king then drank 13 of this liquid mixture. When the king looked away for the third time, the court jester refilled the cup with more water. What percent of this final mixture is red wine?

50%

40%

30%

25%

#### SOLUTION

Solution :B

Method 1:

Let the capacity of glass be 100 ml.

Red WineWaterInitial1000After the first operation8020After the second operation6040After the third operation4060

Method 2:Quantity of red wine left after 3 operations, when the glass originally contains 100% wine from which 15,14 and 13 of total wine are taken out each time and replaced is

=(1−15)(1−14)(1−13)=40%

### Question 6

In how many ways can 80 be written as the difference of squares of two natural numbers?

2

3

4

5

#### SOLUTION

Solution :B

To write a number as a difference of squares of 2 natural numbers, we need to find the 2 factor products of the number, such that both the factors are either even or odd.This is because if we are writing a number N as a difference of squares, then -

N=a2−b2⟹N=(a+b)(a−b). If (a+b, a-b) is a pair of one odd and one even number, 'a' and 'b' can't be natural numbers.

Hence, both the factors need to be either even or odd.

In the case of 80, the possible cases are 80=2×40; =4×20;=8×10.

Thus, there are only three instances where both are even. Hence, the answer is option (b).

### Question 7

If a 3-digit number 'abc' has 2 factors, how many factors does the 6-digit number 'abcabc' have?

16

4

24

8

#### SOLUTION

Solution :A

'abc' has 2 factors.

This means 'abc' is a prime number (Only a prime number can have exactly 2 factors).

Now, 'abcabc' = 'abc'×1001

'abcabc' = 'abc' × 7 × 11 × 13

Since 'abc' is prime we can write 'abcabc' as - p1×71×111×131

No. of factors = (1+1) (1+1) (1+1) (1+1) = 16 factors.

### Question 8

Consider the region formed by the intersections of three semi circles whose diameters, of lengths 3, 4 and 5, from a triangle. What is the area of shaded region bounded by the arcs AB, BC and AC?

6

12

3π

6π

#### SOLUTION

Solution :A

A Radius of semicircles ACB, AC and BC are 2.5, 2 and 1.5 respectively.

Area of Semi circles (ACB), (AC) and (BC) are 6.25π2, 2π and 2.25π2 respectively. And area of

triangle ABC =12 × 4 × 3 = 6

Sum of the area of region A and B = (Area of semicircle (ACB) - Area of triangle (ABC)} = (6.25π2−6)

Area of Shaded region = (Sum of the area of semicircles AC and BC - Sum of the area of region A and B)

Area of Shaded region = (2π+2.25π2) - (6.25π2−6) = (6.25π2) - (6.25π2−6) = 6

Alternate Solution:

As the options are far apart we can easily arrive at the answer by approximation. The shaded

region is approximately half of the areas of semicircles AC and BC = 14(4π+2.25π) = 6(approximately)

### Question 9

According to a survey, at least 65% of the people like BlackBerry, 75% like iPhone and 80% like Samsung. It is known that everyone in the survey likes atleast one of these three. What is the minimum percentage of people who like all three?

10%

15%

20%

25%

#### SOLUTION

Solution :C

Let total number of people = X = 100.

No. of people who like BlackBerry = 65

No. of people who like iPhone = 75

No. of people who like Samsung = 80

Now, we need to find IIImin which represents the minimum percentage of people who like all three.

IIImin = 100 - ((X-65) + (X-75) + (X-80))

Therefore, IIImin = 100 - 35 - 25 - 20 = 20.

Thus, required answer = 20%.

### Question 10

Two clocks are set to show the correct time at 1:00am on a day. One clock loses 3 minutes in an hour and the other clock gains 3 minutes in an hour. Exactly after how many days will both the watches show the correct time? Assume that is a 12 hour clock.

30 days

10 days

15 days

5 days

#### SOLUTION

Solution :A

Every clock shall display the correct time when it is late or fast by hours which are a multiple of 12 .

As per the question, the first clock loses 3 minutes in an hour. Therefore, to lose 12 hours, it will take 12×603=240 hours =10 days.

Also, the second clock gains 2 minutes inan hour. Therefore, to gain 12 hours, it will take : 12×602=360 hours = 15 days.

Therefore, after 30 days (LCM of 15 and 10), both the watches will show the correct time .

### Question 11

Box I contains 5 red and 6 black marbles. Box II contains 4 red and 5 black marbles. One marble is drawn at random from Box I and dropped into Box II. A marble is then drawn at random from Box II and dropped into Box I. A marble is now picked at random from Box I. Find the probability that the last marble drawn from Box I is red.

6231210

7431320

6211210

None of the above

#### SOLUTION

Solution :C

A red marble can be drawn from Box I afterthe two rounds of transfers in the following four ways:

Case 1: R→R→R

Probability = (511×5110×511)

Case 2: R→B→R

Probability = (511×5110×411)

Case 3: B→B→R

Probability = (611×6110×511)

Case 4: B→R→R

Probability = (611×6110×611)

Thus resulting probability = sum of probabilities of individual cases

=(511×5110×511) +(511×5110×411) +(611×6110×511) +(611×6110×611) = 6211210

Therefore, answer option (c) is the correct answer choice.

### Question 12

A rectangle is divided into four rectangles, and their respective areas are shown in the figure. (Figure is not drawn to scale). What is the missing area 'X' of the fourth rectangle?

25205x

5sq units

2sq units

4sq units

2.5 sq units

#### SOLUTION

Solution :C

Approach 1:

You can directly mark the answer as 25 × x = 20 × 5 ⇒ x=4

Proof:

Observe that

ac = 25 sq units ....(1)

bc= 20 sq units ... (2)

ad = 5 sq units ... (3)

Dividing (1) by (3), we get: cd = 5 units ... (4)

Dividing (1) by (2) we get ab =5/4 units ... (5)

St Multiplying (4) and (5) we get:

Sit acbd = 254 sq units

Since its already given that ac = 25 sq units, therefore the missing area 'X' =4 square units.

Approach 2:

Use the simple number approach to break up the values in each cell as shown below

thus x=4x1=4

### Question 13

Each of the letters A, H, I, M, O, T, U, V, W, X, Y appears same when looked at in the mirror. (No lateral inversion). They are hence called symmetric letters. Other letters in the alphabet are asymmetric letters. How many 4 letter computer passwords can be formed (no repetition allowed) with at least one symmetric letter?

11770

12540

326040

13585

#### SOLUTION

Solution :C

Method 1

There are four possible cases that satisfy the condition of forming 4-letter passwords with at least one symmetric letter.

Case 1: 1 symmetric letter + 3 unsymmetric letters

Case 2: 2 symmetric letters + 2 unsymmetric letters

Case 3: 3 symmetric letters + 1 unsymmetric letter

Case 4: All 4 symmetric letters

= {(11C1 × 15C3) + (11C2 × 15C2) + (11C3 × 15C1) + (11C4) }

=5005 + 5775 + 2475 + 330

=13585 × 4! = 326040

Method 2

Total number of cases- cases where there are no symmetric letters26C4 -15C4 = 13585 × 4! = 326040

### Question 14

Two cars start at the same time from towns A and B, and travel with constant but diﬀerent speeds towards each other. They pass each other at 70 km from town A, and continue to the opposite town, where they turn around without stopping. The cars pass each other again 40 km from town B. The distance between the two towns is

165

170

175

180

#### SOLUTION

Solution :B

Method 1:

Assume that the distance between the towns is x km. The first time the cars pass each other, the first car (from town A) has gone 70 km and the other has gone

(x-70)km. These distance were covered in the same time, so the ratio of the speeds of the cars is 70x−70.

The next time they meet, the first car has travelled (x+40) km and the second car has travelled (2x-40) km.Thus the ratio of the speeds is also equal to (x+40)2x−40. We can equate the two ratios to get 70x−70=(x+40)2x−40 , so 70(2x-40)=(x+40)(x-70). This gives 140x−2800=x2−30x−2800, so, x(x-170) =0. The towns are 170 km apart.

Method 2:Reverse Gear:

Go from answer options,

Since the time is constant in both cases, the distance ratios should also remain the same. Assuming that option (b) is correct

70100=210300

Hence,Distance Ratio 1 (first Meeting =Distance Ratio(Second Meeting) Our Assumption is correct. Answer os option (b)).

### Question 15

Find the area of the shaded region, such that square is formed by using the lines x=±1, y=±1.

4−π2

4−π4

4 - π

2(4 - π)

#### SOLUTION

Solution :A

From the figure, it is evident the required area = Area of Square−Area of Circle4×2

Therefore, required area = [(2)2−π(1)2]4×2=4−π2

Shortcut:4-pi-2 method

This is the most straightforward example of the circle inside a square pattern.

Therefore, we know that Area of square: Area of Circle = 4:π

Thus,required area = 4−π2

### Question 16

Two rugby teams (thirty players) have been selected to play for their school. Five of the players speak Spanish, Afrikaans and English. Nine of them speak only Spanish and English. Twenty speak Afrikaans, of which twelve also speak Spanish. Eighteen speak English. No one speaks only Spanish. How many players speak only English?

5

3

2

1

#### SOLUTION

Solution :D

From the data given in the question, we can draw a venn diagram as follows

Since N=30, the sum of the shaded region is 10. Hence 9+x=10 & x=1. Answer is option (d)

### Question 17

A function f(x) is defined as:

f(x) = x15- 2013x14 + 2013x13- 2013x12 + .... - 2013x2 + 2013x. Find the value of f(2012).

2011

2012

2013

2014

#### SOLUTION

Solution :B

Method 1- Conventional Approach

f(x) = x15- 2013{x14 - x15 + x12 - ......... + x2 - x} or

f(x) = x15- 2013{-x + x2 - x3 + ........ - x13 + x14}

f(x) = x15 - (2013x(x14−1)(x+1))

Now, Put x = 2012

f(2012) = (2012)15 - (2013.2012(201214−1)(2012+1))

f(2012) = (2012)15 - 2012(2012^{14} - 1)

f(2012) = 2012.

Method 2- Assumption

Assume 2012 = n and 2013 = (n+ 1). Put x=1. Now replace the options in terms of "n" as

a) (n-1) b) n c) (n + 1) d) (n + 2)

f(1)=1 - 2 + 2 - 2 + 2 - 2 + 2 - ......................-2 + 2

f(1) = 1. Options (a),(c) & (d) can be eliminated as only option (b) gives us 1.

### Question 18

In the diagram below, how many distinct paths are there from the start to the finish, moving from one adjacent dot to the next either to the right, down, or diagonally down to the right?

362

372

400

370

#### SOLUTION

Solution :B

For each dot in the diagram, we can count the number of paths from the start to it by adding the number of ways to get to the dots to the left of it, above it, and above and to the left of it, starting from the topmost leftmost dot (using a simple logic that the value of each dot is the sum of the values of the incoming dots).

This yields the following numbers of paths:

### Question 19

What is the distance between the points of tangency of a common tangent AB to two circles I and II which are externally tangent to each other. The radii of the two circles are 9 and 16 units.

24

22

27

17√3

#### SOLUTION

Solution :A

From the diagram above, AC1=16

and BC2=9.

Also C1 - C2=25.

PC1 = 7 (16-9)

AB=PC2

C1 - P - C2 is a right angled triangle

(C1−C2)2 = PC12+PC22

Thus, PC2 =√625−49

= √576 = 24

### Question 20

Let, (x) = Least integer greater than or equal to x

[x] = Greatest integer less than or equal to x

|x| = absolute value of x,

Which of the following always holds good if x < 0?

[|x|] = | [x ] |

[|x|]<|[x]|

[|x|]>|[x]|

|x|<[x]

#### SOLUTION

Solution :B

The right answer is [|x|]<|[x]|.

Let x = n + f where n is its integral part & f is its fractional part.

Hence, | x | = n + f if x is positive and | x | = - n - f if x is negative.

If x is - ve, then | x | = -n -f .

Hence,[ | x | ] = -n and | [x] | is always non negative.

Hence the alternative (2) holds good.

Method 2- Using assumption

Take a value say x= -1.4 (you can take anything!)

Thus, | - 1.4| = 1.4 and [1.4] = 1

Also, [ -1.4] = -2 and |-2| = 2

Glance at the answer options; the only one satisfying this assumption is [|x|]<|[x]|.

### Question 21

How many positive even integers less than 50 can be written as a sum of three consecutive positive integers?

7

4

10

8

#### SOLUTION

Solution :D

The sum of even, odd, even is odd and the sum of odd, even, odd is even. To get an even sum, the three consecutive integers are odd, even, and odd (or) should start with odd. Hence we have 1 + 2+ 3 = 6 which is 6 less than 3 + 4 + 5 = 12 since 3 is 2 more than 1, 4 is 2 more than 2 and 5 is 2 more than 3. Therefore 5 + 6 + 7 is 6 more than 3 + 4 + 5. Hence we get all multiples of 6 less than 50. They are 6,12,18,24,30,36,42,48.

Hence option (d) is the correct choice.

Alternate Solution:

Let the three consecutive integers be (a-1), (a), (a+1).

(a - 1) + a + (a + 1) = 3a = 50. So all even multiples of 3 which are less than 50, can be written in the form of 3 consecutive integers.

### Question 22

In a group of people, there are 13 who like apples, 9 who like blueberries, 15 who like cantaloupe,and 6 who like dates. (A person can like more than 1 kind of fruit.) Each person who likes blueberries also likes exactly one of apples and cantaloupe. Each person who likes cantaloupe also likes exactly one of blueberries and dates. Find the minimum possible number of people in the group

15

22

28

19

#### SOLUTION

Solution :B

Everyone who likes cantaloupe likes exactly one of blueberries and dates.

However, there are 15 people who like cantaloupe, 9 who like blueberries, and 6 who like dates. Thus, everyone who likes blueberries or dates must also like cantaloupes (because if any of them didn't, we would end up with less than 15 people who like cantaloupe).

Since everyone who likes blueberries likes cantaloupes, none of them can like apples.

However, the 6 people who like both cantaloupe and dates can also like apples. So, we could have a group where 7 people like apples alone, 9 like blueberries and cantaloupe, and 6 like apples, cantaloupe, and dates. This gives 22 people in the group, which is the minimum possible.

### Question 23

A sphere is inscribed in a cube that has a surface area of 24 m2. A second cube is then inscribed within the sphere. What is the surface area in square metres of the inner cube?

6

8

4

12

None of these

#### SOLUTION

Solution :B

Given that surface area of the original cube = 24 units.

∴ Length of each side of the cube =√246= 2 units.

The radius of the biggest possible sphere that can be kept inside the cube of each side 2 units.

= 22 = 1 unit.

Now, the diagonal of the second cube inscribed in the sphere = Diameter of sphere.

⇒√3 (length of each side of second cube) = 2 units.

⇒Length of each side of second cube =2√3 unit

∴ surface area of the inner cube = 6 x(2√3)2

= 6×43= 8 units

### Question 24

Given a set of n rays in a plane, define a reversal as the operation of reversing precisely one ray and obtaining a new set of rays. If all the rays are reversed after 42 operations, then n can be

21

23

41

24

#### SOLUTION

Solution :D

For a ray to get back to its original orientation, it requires an even no. of operations. Since, all the rays get reversed after 42 operations and if there are k rays ⇒ 2n+k = 42 ⇒ k has to be even. Hence, choice (d) is the right answer

### Question 25

Find the sum of 1.1! + 2.2! + 3.3! + ... + (n-1)(n-1)! + n. n!.

n!

(n - 1)!

(n + 1)! - 1

( n + 1)!

#### SOLUTION

Solution :C

Put n = 2

Given expression = 5

Option (a) = 2

Option (b) = 1

Option (c) = 5

Option (d) = 6Hence option (c)

### Question 26

f(x) = |x|. If f(x) is reflected with respect to the line x -2 =0 to get g(x), then which of the following is g(x)?

g(x+4) = |x|

g(x-4) = |x|

g(x) = |x| - 2

g(x) = |x| + 2

#### SOLUTION

Solution :A

Option (a)

g(4) = 0 also f(0) = 0. So, g(4) = f(0)

Out of the given options, only (a) satisfies g(4) = f(0) when you put x = 0 Hence, the answer is

option (a)

### Question 27

While ordering these four numbers from least to greatest: 536, 1051, 1735, 3117, which is the 3rd number?

536

1051

1735

3117

#### SOLUTION

Solution :C

1051>951=3102=2734>1735>1635=3228>3117>2518=536.

So the ordering is 536<3117<1735<1051.

### Question 28

A trader used to make 5% profit on an item selling at usual marked price. One day, he trebled the marked price of the item and finally offered a discount of

30%. Find the percentage profit he made on the item that day.

120.50%

100%

99.5%

94.5%

#### SOLUTION

Solution :A

Let the earlier cost price of the item = Rs.100

i.e. Earlier marked price = Rs.105.

On that day, 30% discount is offered on Rs. 3 × 105 = Rs.315

Thus, new selling price = Rs.220.50New Profit percentage = 120.50%

### Question 29

A trader used to make 5% profit on an item selling at usual marked price. One day, he trebled the marked price of the item and finally offered a discount of 30%. Find the percentage profit he made on the item that day.

Rs.8

Rs.9

Rs.10

Rs.12

#### SOLUTION

Solution :D

Method 1:Conventional approach

Cost of a tubelight = Rs. t and cost of a bulb = Rs. b

t + b = 52

Cost of bulb drops by 20% and cost of tubelight increases by 50%

Therefore, 1.5t + 0.8b = 50

Solving the two equations in 't' and 'b',

t + b = 52

1.5t + 0.8b = 50

Multiplying first equation by 4 and second equation by 5,

4t + 4b = 208

7.5t + 4b = 250

Subtracting,

3.5t = 42

Hence, t = 12

Thus, cost of a tubelight = Rs. 12

Shortcut!- Reverse Gear

Checking for option (c)Tubelight1015Bulb42335248

From this we can deduce that the cost of the tubelight is very close to our assumption. We can also directly mark option (d) as the correct answer without verifying because with a Rs.10 tubelight we are getting the new cost as Rs.48 ; thus the cost of the tubelight has to be slightly more than Rs.10 to get the cost of the package as Rs.50 !

### Question 30

The pair of two-digit numbers 12 and 63 has the interesting property that if the product expression 12×63 is read in reverse order 36×21, the two product expressions are equal, i.e. 12×63=36×21. How many pairs of two-digit numbers with this property are there in which one of the numbers in the pair is greater than 20 and less than 30, and the units' digit of the other number in the pair is an odd prime?

#### SOLUTION

Solution :(20 + x)(10y + p) = (10p + y)(10x + 2) ⇒ px = 2y. For p = 5 or 7 ⇒ y = p and x = 2, and for p = 3 ⇒ y = 3, 6 or 9 and x = 2, 4 or 6 respectively.

### Question 31

Find the sum of all the integers N>1 such that the each prime factor of N is either 2, 3 or 7 and N is not divisible by any perfect cube greater than1.

#### SOLUTION

Solution :The number will be of the form (2x)×(3y)×(7x).

And 0≤x,y,z≤2 (Since N is not divisible by any perfect cube greater than 1).

Hence sum of all possible combination of numbers is: (1+2+4)×(1+3+9)×(1+7+49)−1=5187−1=5186

### Question 32

A book, comprising of A3 size sheets bound in the centre, form a book with 200 A3 size pages numbered one to 200 back to back. The middle A3 sheet is then pulled out from the book. What is the resulting sum of all the page numbers still remaining in the book?

#### SOLUTION

Solution :The initial sum of all pages in the book =200×2012=20100.

If we try to visualize the binding of the book, it would consist of 100 leaves, printed back to back so that the total number of pages in the book = 200.

If we pull out the middle sheet from the book, pages 99, 100, 101 and 102 would get pulled out.

Therefore, resulting sum of all the page numbers still remaining in the book = 20100 - (99+100+101+102) = 19698

### Question 33

The painting of a hotel can be done by 10 painters in 5 days. But, in reality there were few dabblers in the group of 10 painters and therefore the work took a little over 6 days for completion. A painter worked twice as fast as the dabbler and used 25% less paint than a dabbler used. If 5 litres extra paint was used than expected, then the total paint used in painting the hotel was (in litres)

#### SOLUTION

Solution :Option (d)

Let the number of painters be x, then dabblers is 10 - x

Now x50+(10−x)100= slightly less than 12

(2x+10−x)100= slightly less than 16

x+10 = slightly less than 1006

x= slightly less than 6.67

Since x has to be integer nearest integer is 6.

That means the number of dabblers = 4.

Let p be the amount of paint used by each dabbler, so by each painter = .75p

Extra paint used =.25p×4= 5 litres

So p = 5 litres.

Dabblers will be using 5×4 = 20 litres of paint. Now as the painters are working at twice the efficiency, they will use twice the amount of 75% of paint used by dabbler in the same time.

i.e. paint used by each painter =2×(.75)×5 = 7.5 litres.

Hence paint used by 6 painters =7.5×6= 45 litresSo total paint used = 45 + 20 = 65litres

### Question 34

Find the sum of every even positive integer less than 233 not divisible by 10

#### SOLUTION

Solution :We find the sum of all positive even integers less than 233 and then subtract all the positive integers less than 233 that are divisible by 10.

2+4+....+232 = 2(1+2+. . . +116) = 116 .117 = 13572.

The sum of all positive integers less than 233 that are divisible by 10 is 10+20+. . .+230 = 10(1+2+. . .+23) = 2760. Then our answer is 13572-2760 = 10812.