# Arithmetic Questions for CAT Set-3 PDF

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## Arithmetic Questions for CAT Set-3 PDF

Download important CAT Arithmetic Set-3 Questions PDF based on previously asked questions in CAT exam. Practice Arithmetic Set-3 Questions PDF for CAT exam.

Question 1: If R = $(30^{65}-29^{65})/(30^{64}+29^{64})$ ,then

a) $0<R\leq0.1$

b) $0.1<R\leq0.5$

c) $0.5<R\leq1.0$

d) $R>1.0$

Question 2: If a number 774958A96B is to be divisible by 8 and 9, the respective values of A and B will be

a) 7 and 8

b) 8 and 0

c) 5 and 8

d) None of these

Question 3: 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) 0

b) 1

c) (1/2)*n

d) (n+1)/2n

e) 2008

Question 4: A college has raised 75% of the amount it needs for a new building by receiving an average donation of Rs. 600 from the people already solicited. The people already solicited represent 60% of the people the college will ask for donations. If the college is to raise exactly the amount needed for the new building, what should be the average donation from the remaining people to be solicited?

a) Rs. 300

b) Rs. 250

c) Rs. 400

d) 500

Question 5: A student gets an aggregate of 60% marks in five subjects in the ratio 10 : 9 : 8 : 7 : 6. If the passing marks are 50% of the maximum marks and each subject has the same maximum marks, in how many subjects did he pass the examination?

a) 2

b) 3

c) 4

d) 5

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Question 6: I have one-rupee coins, 50-paisa coins and 25-paisa coins. The number of coins are in the ratio 2.5 : 3 : 4. If the total amount with me is Rs. 210, find the number of one-rupee coins.

a) 90

b) 85

c) 100

d) 105

Question 7: Fresh grapes contain 90% water by weight while dried grapes contain 20% water by weight and the remaining proportion being pulp. What is the weight of dry grapes available from 20 kg of fresh grapes?

a) 2 kg

b) 2.4 kg

c) 2.5 kg

d) None of these

Question 8: The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is

a) 3 : 10

b) 1 : 3

c) 1 : 4

d) 2 : 5

Question 9: I sold two watches for Rs. 300 each, one at the loss of 10% and the other at the profit of 10%. What is the percentage of loss(-) or profit(+) that resulted from the transaction?

a) (+)10

b) (-)1

c) (+)1

d) (-)10

Question 10: How many pairs of positive integers m, n satisfy 1/m + 4/n = 1/12 , where n is an odd integer less than 60?

a) 6

b) 4

c) 7

d) 5

e) 3

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$\frac{(30^{65}-29^{65})}{(30^{64}+29^{64})} = ((30-29)*\frac{(30^{64}+30^{63}*29+….+29^{64})}{(30^{64}+29^{64})}$ , which is greater than 1 . Hence option D.

According to the divisible rule of 9, the sum of all digits should be divisible by 9.
i.e. 55+A+B = 9k
So sum can be either 63 or 72.
For 63, A+B should be 8.
In given options, option B has values of A and B whose sum is 8 and by putting them we are having a number which divisible by both 9 and 8.
Hence answer will be B.

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 = $n^2 + 2n$/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

Let there be total 100 people whom the college will ask for donation. Out of these 60 people have already given average donation of 600 Rs. Thus total amount generated by 60 people is 36000. This is 75% of total amount required . so the amount remaining is 12000 which should be generated from remaining 40 people. So average amount needed is 12000/40 = 300

Let’s say he scored marks as $10x,9x,8x,7x,6x$ or total of $40x$ which is 60% of total maximum marks(T).

$\frac{T \times 60}{100}=40x$

So T (total maximum marks)=$\frac{400x}{6}$

Or Individual max. marks = $\frac{T}{5}=\frac{80x}{6}$

Passing marks =50% of individual max. marks =$\frac{40x}{6}=6.66x$

Hence he scored more than passing marks in four subjects as $10x,9x,8x$ and $7x$

and failed in one subject as scoring $6x$ marks which is less than passing marks of $6.66x$

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Let’s say number of coins are 2.5x , 3x and 4x
So total amount will be = 2.5x + 3x(0.5) + 4x(0.25) = 210
So x = 42
And number of 1 rs. coins = 2.5x = 105

Fresh grapes contain 90% water so water in 20kg of fresh pulp = (90/100)x20= 18kg.

In 20kg fresh grapes, the weight of water is 18kg and the weight of pulp is 2kg.

The concept that we apply in this question is that the weight of pulp will remain the same in both dry and fresh grapes. If this grape is dried, the water content will change but pulp content will remain the same.

Suppose the weight of the dry grapes be D.

80% of the weight of dry grapes = weight of the pulp = 2 kg

(80/100) x D =2 kg.

D = 2.5 kg

Let ‘a’, ‘b’ and ‘c’ be the concentration of salt in solutions A, B and C respectively.

It is given that three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%.

$\Rightarrow$ $\dfrac{a+2b+3c}{1+2+3} = 20$

$\Rightarrow$ $a+2b+3c = 120$ … (1)

If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%.

$\Rightarrow$ $\dfrac{3a+2b+c}{1+2+3} = 30$

$\Rightarrow$ $3a+2b+c = 180$ … (2)

From equation (1) and (2), we can say that

$\Rightarrow$ $b+2c = 45$

$\Rightarrow$ $b = 45 – 2c$

Also, on subtracting (1) from (2), we get

$a – c = 30$

$\Rightarrow$ $a = 30 + c$

In solution D, B and C are mixed in the ratio 2 : 7

So, the concentration of salt in D = $\dfrac{2b + 7c}{9}$ = $\dfrac{90 – 4c + 7c}{9}$ = $\dfrac{90 + 3c}{9}$

Required ratio = $\dfrac{90 + 3c}{9a}$ = $\dfrac{90 + 3c}{9 (30 + c)}$ = $1 : 3$

Hence, option B is the correct answer.

Selling price of first watch = 300
Profit = 10%
cost price = $\frac{300}{1.1}$
Selling price of second watch = 300
Loss = 10%
cost price = $\frac{300}{0.9}$

Total selling price of transaction= 600
Total cost price of transaction = $300(\frac{10}{11} + \frac{10}{9}) = 600 (\frac{100}{99})$
Loss = $600 (\frac{100}{99} – 1)$
%loss = $(600 (\frac{100}{99} – 1)) \div (600(\frac{100}{99})) \times 100 = 1$