0
955

# CAT Arithmetic Questions PDF [Most Important]

Arithmetic topic is the most important topic in the CAT Quantitative Ability (QA) Section. It is very essential that you know the basics of the CAT Arithmetic section well and practise the questions well. Also, do check out all the arithmetic questions from the CAT Previous Papers with detailed video solutions. This article will look into some Arithmetic Questions for the CAT Exam. If you want to practice these important Arithmetic questions, you can download the PDF, which is completely Free.

Question 1:Â Consider the set S = {2, 3, 4, …., 2n+1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X – Y ?

a)Â 0

b)Â 1

c)Â (1/2)*n

d)Â (n+1)/2n

e)Â 2008

Solution:

The odd numbers in the set are 3, 5, 7, …2n+1

Sum of the odd numbers = 3+5+7+…+(2n+1) = $n^2 + 2n$

Average of odd numbers =Â $n^2 + 2n$/n = n+2

Sum of even numbers = 2 + 4 + 6 + … + 2n = 2(1+2+3+…+n) = 2*n*(n+1)/2 = n(n+1)

Average of even numbers = n(n+1)/n = n+1

So, difference between the averages of even and odd numbers = 1

Question 2:Â Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eight-member joint family is nearest to
[CAT 2007]

a)Â 23 years

b)Â 22 years

c)Â 21 years

d)Â 25 years

e)Â 24 years

Solution:

Ten years ago, the total age of the family is 231 years.

Seven years ago, (Just before the death of the first person), the total age of the family would have been 231+8*3 = 231+24 = 255.
This is because, in 3 years, every person in the family would have aged by 3 years,
Total change in age = 231+24 = 255

After the death of one member, the total age is 255-60 =Â 195 years.

Since a child takes birth in the same year, the number of members remain the same i.e. (7+1) = 8

Four years ago, (i.e. 6 years after start date)Â one of the member of age 60 dies,
therefore,Â total age of the family is 195+24-60 = 159 years.

Since a child takes birth in the same year, the number of members remain the same i.e. (7+1) = 8

After 4 more years, the current total age of the family is = 8×4 + 159 = 191 years

The average age is 191/8 = 23.875 years = 24 years (approx)

Alternatively,
Since the number of members is always the same throughout
The 2 older members dropped their age by 60
So, after 10yrs, total age = 231 + 8*10 – 2*60Â = 191
Average age = 191/8 = 23.875 $\simeq$ 24

Question 3:Â Let A and B be two solid spheres such that the surface area of B is 300% higher than the surface area of A. The volume of A is found to be k% lower than the volume of B. The value of k must be
[CAT 2003 leaked]

a)Â 85.5

b)Â 92.5

c)Â 90.5

d)Â 87.5

Solution:

Surface area of sphere A (of radius a) is $4\pi*a^2$
Surface area of sphere B (of radius b) is $4\pi*b^2$
=> $4\pi*a^2$/$4\pi*b^2$ = 1/4 => a:b = 1:2
Their volumes would be in the ratio 1:8
Therefore, k = 7/8 * 100% = 87.5%

Question 4:Â A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?
[CAT 2004]

a)Â 2 : 3

b)Â 1 : 2

c)Â 1 : 3

d)Â 3 : 4

Solution:

After selling 1/4th of the mixture, the remaining quantity of water is 15 liters and milk is 60 liters. So the milkman would add 25 liters of water to the mixture. The total amount of water now is 40 liters and milk is 60 liters. Therefore, the required ratio is 2:3.

Question 5:Â A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 6 : 7 : 8 : 9 : 10. In all papers together, the candidate obtained 60% of the total marks. Then the number of papers in which he got more than 50% marks is

a)Â 2

b)Â 3

c)Â 4

d)Â 5

Solution:

Let the marks in the five papers be 6k, 7k, 8k, 9k and 10k respectively.
So, the total marks in all the 5 papers put together is 40k. This is equal to 60% of the total maximum marks. So, the total maximum marks is 5/3 * 40k
So, the maximum marks in each paper is 5/3 * 40k / 5 = 40k/3 = 13.33k
50% of the maximum marks is 6.67k
So, the number of papers in which the student scored more than 50% is 4

Question 6:Â Karan and Arjun run a 100-meter race, where Karan beats Arjun 10 metres. To do a favour to Arjun, Karan starts 10 metres behind the starting line in a second 100 metre race. They both run at their earlier speeds. Which of the following is true in connection with the second race?

a)Â Karan and Arjun reach the finishing line simultaneously.

b)Â Arjun beats Karan by 1 metre

c)Â Arjun beats Karan by 11 metres.

d)Â Karan beats Arjun by 1 metre.

Solution:

The speeds of Karan and Arjun are in the ratio 10:9. Let the speeds be 10s and 9s.
Time taken by Karan to cover 110 m = 110/10s = 11/s
Time taken by Arjun to cover 100 m = 100/9s = 11.11/s
Therefore, Karan reaches the finish line before Arjun. From the options, the only possible answer is d).

Question 7:Â Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30, 40 and 60 km per hour respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start?

a)Â 3

b)Â 3.5

c)Â 4

d)Â 4.5

e)Â 5

Solution:

Let the distance be D.
Time taken by Arun = D/30
Time taken by Barun = D/40
Now, D/40 = D/30 – 2
=> 3D = 4D – 240
=> D = 240
Therefore time taken by Arun to cover 240 km = 240/30 = 8 hr
Time Kiranmala takes to cover 240 km = 240/60 = 4 hr
So, Kiranmala has to start 4 hours after Arun.

Question 8:Â At his usual rowing rate, Rahul can travel 12 miles downstream in a certain river in 6 hr less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for this 24 miles round trip, the downstream 12 miles would then take only 1 hr less than the upstream 12 miles. What is the speed of the current in miles per hour?

a)Â $\frac{7}{3}$

b)Â $\frac{4}{3}$

c)Â $\frac{5}{3}$

d)Â $\frac{8}{3}$

Solution:

$12/(R – S) = T$
$12/(R + S) = T – 6$
$12/(2R – S) = t$
$12/(2R + S) = t – 1$
=> $12/(R – S) – 12/(R + S) = 6$ and $12/(2R – S) – 12/(2R + S) = 1$
=> $12R + 12S – 12R + 12S = 6R^2 – 6S^2$ and $24R + 12S – 24R + 12S = 4R^2 – S^2$
=> $24S = 6R^2 – 6S^2 and 24S = 4R^2 – S^2$
=> $6R^2 – 6S^2 = 4R^2 – S^2$
=> $2R^2 = 5S^2$
=> $24S = 10S^2 – S^2 = 9S^2$
=> $S = 24/9 = 8/3$

Question 9:Â Shyama and Vyom walk up an escalator (moving stairway). The escalator moves at a constant speed. Shyama takes three steps for every two of Vyomâ€™s steps. Shyama gets to the top of the escalator after having taken 25 steps, while Vyom (because his slower pace lets the escalator do a little more of the work) takes only 20 steps to reach the top. If the escalator were turned off, how many steps would they have to take to walk up?

a)Â 40

b)Â 50

c)Â 60

d)Â 80

Solution:

Let the number of steps on the escalator be x.

So, by the time Shyama covered 25 steps, the escalator moved ‘x-25’ steps.

Hence, the ratio of speeds of Shyama and escalator = 25:(x-25)

Similarly, the ratio of speeds of Vyom and escalator = 20:(x-20)

But the ratio is 3:2

Ratio of speeds of Shyama and Vyom = 25(x-20)/20*(x-25) = 3/2

=> 10(x-20) = 12(x-25)

=> 2x = 100 => x = 50

Question 10:Â Thereâ€™s a lot of work in preparing a birthday dinner. Even after the turkey is in the oven, thereâ€™s still the potatoes and gravy, yams, salad, and cranberries, not to mention setting the table.
Three friends â€” Asit, Arnold and Afzal â€” work together to get all of these chores done. The time it takes them to do the work together is 6 hr less than Asit would have taken working alone, 1 hr less than Arnold would have taken alone, and half the time Afzal would have taken working alone. How long did it take them to do these chores working together?

a)Â 20 min

b)Â 30 min

c)Â 40 min

d)Â 50 min

Solution:

Let the time taken working together be t.
Time taken by Arnold = t+1
Time taken by Asit = t+6
Time taken by Afzal = 2t
Work done by each person in one day =Â $\frac{1}{(t+1)}+\frac{1}{(t+6)}+\frac{1}{2t}$
Total portion of workdone in one day $=\frac{1}{t}$
$\frac{1}{(t+1)}+\frac{1}{(t+6)}+\frac{1}{2t}=\frac{1}{t}$
$\frac{1}{(t+1)}+\frac{1}{(t+6)}=\frac{2-1}{2t}$
$2t+7=\frac{(t+1)\cdot(t+6)}{2t}$
$3t^2-7t+6=0Â \longrightarrow\ t=\frac{2}{3}$or $t=-3$
Therefore total time = $\frac{2}{3}$hours = 40mins

Alternatively,
$\frac{1}{(t+1)}+\frac{1}{(t+6)}+\frac{1}{2t}=\frac{1}{t}$
From the options, if time $= 40$ min, that is, $t = \frac{2}{3}$
LHS =Â $\frac{3}{5} + \frac{3}{20} + \frac{3}{4} = \frac{(12+3+15)}{20} = \frac{30}{20} = \frac{3}{2}$
RHS = $\frac{1}{t}=\frac{3}{2}$
The equation is satisfied only in case of option C
Hence, C is correct

Instructions

Directions for the following two questions: Shabnam is considering three alternatives to invest her surplus cash for a week. She wishes to guarantee maximum returns on her investment. She has three options, each of which can be utilized fully or partially in conjunction with others.

Option A: Invest in a public sector bank. It promises a return of +0.10%.

Option B: Invest in mutual funds of ABC Ltd. A rise in the stock market will result in a return of +5%, while a fall will entail a return of â€“ 3%.

Option C: Invest in mutual funds of CBA Ltd. A rise in the stock market will result in a return of â€“ 2.5%, while a fall will entail a return of + 2%.

a)Â 0.25%

b)Â 0.10%

c)Â 0.20%

d)Â 0.15%

e)Â 0.30%

Solution:

Let a, b and c be the percentages of amountÂ invested in options A, B and C respectively => a + b + c = 100

Return attained if there is a rise in the stock market => 0.001a + 0.05b – 0.025c

Return attained if there is a fall in the stock market => 0.001a –Â 0.03b +Â 0.02c

Maximum guaranteed return is attained when both are equal because it is indifferent to rise and fall in the market.

0.001a + 0.05b – 0.025c =Â 0.001a –Â 0.03b +Â 0.02c

=> 0.08b = 0.045c => 16b = 9c

Let’s put the values for a, b and c that satisfy the above equation.

b = 9, c = 16,Â a = 75 => return = 0.125

b = 18, c = 32, a = 50 => return = 0.15

b = 27, c = 48, a = 25 => return = 0.175

b = 36, c = 64, a = 0 => return = 0.2

Hence, the maximum guaranteed returnÂ is 0.2%

a)Â 100% in option A

b)Â 36% in option B and 64% in option C

c)Â 64% in option B and 36% in option C

d)Â 1/3 in each of the three options

e)Â 30% in option A, 32% in option B and 38% in option C

Solution:

Let a, b and c be the percentages of amountÂ invested in options A, B and C respectively => a + b + c = 100

Return attained if there is a rise in the stock market => 0.001a + 0.05b – 0.025c

Return attained if there is a fall in the stock market => 0.001a –Â 0.03b +Â 0.02c

Maximum guaranteed return is attained when both are equal because it is indifferent to rise and fall in the market.

0.001a + 0.05b – 0.025c =Â 0.001a –Â 0.03b +Â 0.02c

=> 0.08b = 0.045c => 16b = 9c

Let’s put the values for a, b and c that satisfy the above equation.

b = 9, c = 16,Â a = 75 => return = 0.125

b = 18, c = 32, a = 50 => return = 0.15

b = 27, c = 48, a = 25 => return = 0.175

b = 36, c = 64, a = 0 => return = 0.2

Hence, the maximum guaranteed returnÂ is 0.2% and it is attained when 36% is invested in option B and 64% is invested in option C.

Question 13:Â The owner of an art shop conducts his business in the following manner: every once in a while he raises his prices by X%, then a while later he reduces all the new prices by X%. After one such updown cycle, the price of a painting decreased by Rs. 441. After a second up-down cycle the painting was sold for Rs. 1,944.81. What was the original price of the painting?

a)Â Rs. 2,756.25

b)Â Rs. 2,256.25

c)Â Rs. 2,500

d)Â Rs. 2,000

Solution:

Let the price of the painting be P
One cycle of price increase and decrease reduces the price by $x^2/100 * P = 441$
Let the new price be N =>Â $P – x^2/100 * P = N$
Price after the second cycle =Â $N – x^2/100 * N$Â = 1944.81
=>Â $(P – x^2/100 * P)(1 – x^2/100) = 1944.81$
=>Â $(P – 441)(1 – 441/P) = 1944.81$
=>Â $P – 441 – 441 + 441^2/P = 1944.81$
=>Â $P^2 – (882 + 1944.81)P + 441^2 = 0$
=>Â $P^2 – 2826.81P + 441^2 = 0$
From the options, the value 2756.25 satisfies the equation.
So, the price of the article is Rs 2756.25

Question 14:Â In a stockpile of products produced by three machines M1, M2 and M3, 40% and 30% were manufactured by M1 and M2 respectively. 3% of the products of M1 are defective, 1% of products of M2 defective, while 95% of the products of M3 are not defective. What is the percentage of defective in the stockpile?

a)Â 3%

b)Â 5%

c)Â 2.5%

d)Â 4%

Solution:

Let’s say total products maufactured by M1, M2 and M3 are 100.

So M1 produced 40, M2 produced 30 and M3 produced 30

Defective pieces for M1 = $\frac{120}{100}$

Defective pieces for M2 = $\frac{30}{100}$

Defective pieces for M3 = $\frac{150}{100}$

So total defective pieces are $\frac{150+30+120}{100}$ = $\frac{300}{100}$ = 3% of total products.

Question 15:Â Gopal went to a fruit market with certain amount of money. With this money he can buy either 50 oranges or 40 mangoes. He retains 10% of the money for taxi fare. If he buys 20 mangoes, then the number of oranges he can buy is

a)Â 25

b)Â 18

c)Â 20

d)Â None of these

Let’s say total money was $x$ rs.
So cost price of 40 mango will be = $x$ ;
Hence cost price of 20 mangoes will be = $\frac{x}{2}$
Taxi fare = $\frac{10x}{100}$
Total expense =Â $\frac{x}{2}$ +Â Â $\frac{10x}{100}$ = $\frac{6x}{10}$
Remaining money =$\frac{4x}{10}$
Cost price of 1 orange will be = $\frac{x}{50}$
Hence inÂ $\frac{4x}{10}$ rs. 20 oranges can be purchased.