# Free Quant Practice Test - CAT

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

A.

60

B.

120

C.

64

D.

128

#### SOLUTION

Solution : C

510=2×3×5×17.
Use the concept of Euler's theorem:
510(112)(113)(115)(1117)=128
Number of pairs=1282=64.

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:

A.

N

B.

8N+114

C.

N(N+1)6

D.

8N+112

#### 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)2n2=n
Total number of boxes = 1+2+3+....+2n= n(2n+1) = N
Therefore, 2n2+nN=0
n=1±1+8N4
Since n cannot be negative, n=(8N+1)14.

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.

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

A.

190

B.

172

C.

96

D.

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=100

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.

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?

A.

50%

B.

40%

C.

30%

D.

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
=(115)(114)(113)=40%

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

A.

2

B.

3

C.

4

D.

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=a2b2N=(a+b)(ab). 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).

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

A.

16

B.

4

C.

24

D.

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.

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?

A.

6

B.

12

C.

3π

D.

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)} = (6.25π26)
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) - (6.25π26) = (6.25π2) - (6.25π26) = 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)

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?

A.

10%

B.

15%

C.

20%

D.

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.

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.

A.

30 days

B.

10 days

C.

15 days

D.

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 .

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.

A.

6231210

B.

7431320

C.

6211210

D.

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: RRR
Probability = (511×5110×511)
Case 2: RBR
Probability = (511×5110×411)
Case 3: BBR
Probability = (611×6110×511)
Case 4: BRR
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

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

A.

5sq units

B.

2sq units

C.

4sq units

D.

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

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?

A.

11770

B.

12540

C.

326040

D.

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 letters

26C4 -15C4 = 13585 × 4! = 326040

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

A.

165

B.

170

C.

175

D.

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 70x70.
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)2x40. We can equate the two ratios to get 70x70=(x+40)2x40 , so 70(2x-40)=(x+40)(x-70). This gives 140x2800=x230x2800, so, x(x-170) =0. The towns are 170 km apart.

Method 2: Reverse Gear:
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)).

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

A.

4π2

B.

4π4

C.

4 - π

D.

2(4 - π)

#### SOLUTION

Solution : A

From the figure, it is evident the required area = Area of SquareArea of Circle4×2
Therefore, required area = [(2)2π(1)2]4×2=4π2

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 = 4π2

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?

A.

5

B.

3

C.

2

D.

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)

A function f(x) is defined as:
f(x) = x15- 2013x14 + 2013x13- 2013x12 + .... - 2013x2 + 2013x. Find the value of f(2012).

A.

2011

B.

2012

C.

2013

D.

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(x141)(x+1))
Now, Put x = 2012
f(2012) = (2012)15 - (2013.2012(2012141)(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.

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?

A.

362

B.

372

C.

400

D.

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:

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.

A.

24

B.

22

C.

27

D.

173

#### 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
(C1C2)2 = PC12+PC22
Thus, PC2 =62549
= 576 = 24

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?

A.

[|x|] = | [x ] |

B.

[|x|]<|[x]|

C.

[|x|]>|[x]|

D.

|x|<[x]

#### SOLUTION

Solution : B

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]|.

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

A.

7

B.

4

C.

10

D.

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.

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

A.

15

B.

22

C.

28

D.

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.

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?

A.

6

B.

8

C.

4

D.

12

E.

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 =23 unit
surface area of the inner cube = 6 x(23)2
= 6×43= 8 units

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

A.

21

B.

23

C.

41

D.

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

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

A.

n!

B.

(n - 1)!

C.

(n + 1)! - 1

D.

( n + 1)!

#### SOLUTION

Solution : C

Put n = 2
Given expression = 5
Option (a) = 2
Option (b) = 1
Option (c) = 5
Option (d) = 6

Hence option (c)

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)?

A.

g(x+4) = |x|

B.

g(x-4) = |x|

C.

g(x) = |x| - 2

D.

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)

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

A.

536

B.

1051

C.

1735

D.

3117

#### SOLUTION

Solution : C

1051>951=3102=2734>1735>1635=3228>3117>2518=536.
So the ordering is 536<3117<1735<1051.

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.

A.

120.50%

B.

100%

C.

99.5%

D.

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.220.50

New Profit percentage = 120.50%

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.

A.

Rs.8

B.

Rs.9

C.

Rs.10

D.

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 !

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.

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 0x,y,z2 (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=51871=5186

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

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+(10x)100= slightly less than 12
(2x+10x)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 litres

So total paint used = 45 + 20 = 65litres

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.