# Free Arithmetic Miscellaneous - 02 Practice Test - CAT

### Question 1

Arun invests in bonds worth Rs. 40000 and he receives a 15% interest per annum. How much money should he invest in bonds that give 10% per annum so that he gains overall 12% per annum?

#### SOLUTION

Solution :C

Using alligation

He should invest in the ratio 2:3

He should invest 60000

### Question 2

In a company, there are initially ‘n’ employees. First p% employees were hired and after a month q% of the employees were fired. This again made the number of employees in the company as ‘n’. Find the value of p – q.

#### SOLUTION

Solution :A

n(1+p100)(1−q100)=n/n(1+p100−q100−pq1002)/n1+p100−q100−pq1002=1p100−q100=pq1002⇒p−q=pq100

### Question 3

When the air-conditioner is on, a typist can type X pages per hour. However, when the air-conditioner is off, she can type at 65% of the earlier efficiency (when the air-conditioner is on). How many hours would she take to type out 575 pages when the air-conditioner is off?

#### SOLUTION

Solution :B

Let X=100, Then when AC is off, speed = 65 pages per hour.

Thus to type 575 pages, she takes 57565=8.84 pages per hour. Answer is option (b

Alternatively:

When the air-conditioner is off, 0.65X pages are typed per hour. 575 pages will be typed in5750.65X=884.6X−1

### Question 4

In an automated plant assembly line, the rate of rejection of components was 10% on July 1st and 6% on July 2nd. The combined rate of rejection for the two days was 9% . The ratio of production volumes on July 1st and on July 2nd is

#### SOLUTION

Solution :C

Use alligation

Answer is option (c)

### Question 5

In a factory, a part of the 1500 employees reported for the work while others were on leave. The next day, 4% of the workers who were on leave on the first day reported for the work and 6% of the workers who came on the 1st day joined the strike. If the number of workers on leave on both days were the same, then how many workers were on leave?

#### SOLUTION

Solution :D

Let workers on leave: present workers = x: y.

0.04x come to work and 0.06y take leave on the second day. As there will be no difference in value of x and y after this exchange: 0.04x = 0.06y

x: y = 6: 4 or 3: 2.Number of workers on leave =(35)×1500=900

Hence option (d)

Alternatively, go from answer options

2nd day number joining and number on leave should be the same

This happens only when the number of leave = 900

2nd day = 4% of 900−6% of 600=0 . Option (d)

### Question 6

When Sunil and Tanuj went for shopping initially Sunil had twice the money than Tanuj. They together bought things amounting to Rs.250. Out of which Tanuj’s share was 60%. At the end Sunil was left with thrice the amount that Tanuj had. What was the amount with Tanuj at the beginning

#### SOLUTION

Solution :Initial money with Sunil = s and Tanuj = t

Purchases made by them = Rs. 250

Tanuj’s purchase = 60% of Rs. 250 = Rs. 150

Sunil’s purchase = Rs. 250 – Rs. 150 = Rs. 100

So, money left in the end (s – 100) and (t – 150)

Given s = 2t and (s – 100) = 3 (t – 150)

Solving we get t = 350.

### Question 7

There is an alloy (A) of silver and copper. A certain weight of this alloy is mixed with 15 kg of pure Silver and melted. The new alloy (B) contains 90% of silver. If the alloy (A) is mixed with 10kg of a 90% silver alloy, the new alloy (C) is found to contain 84% silver. Find the percentage of silver in (A).

#### SOLUTION

Solution :A

Use alligation and then the answer options

Let the percentage of silver in alloy A be a

Now substitute values from answer options such that all the criteria are satisfied. Answer is option (a) as shown below

### Question 8

Raman plans to sell his goods at a loss of 8% but uses weights of 900 grams in place of a kg weight. Find his real loss or gain percent?

3%

20%

10%

2.22%

#### SOLUTION

Solution :D

Raman plans to sell his goods at a loss of 8%. Therefore, Selling Price =(100−8)% of CP = 0.92 CP. But, when he uses weights that measure only 900 grams while he claims to measure 1 kg.Hence,

CP of 900gms =0.90×Original CP

So, he is selling goods worth 0.90 CP at 0.92 CPTherefore, he makes a profit of 0.02 CP on his cost of 0.9 CP profit %=2.22%

### Question 9

Ravi took a loan of Rs. 1 Lakh from SBI at the rate of 6% per annum of simple interest for 10 years. After some time, the rate of interest dropped down to 4% and as a result, Ravi paid only Rs. 1.48 lakhs at the end of the loan period. Number of years after which the rate of interest was decreased to 4% is

#### SOLUTION

Solution :He paid Rs. 48,000 interest in 10 years.48,0001,00,000×100=48% in 10 years. So, 4.8% in a year.

After 4 years the rate of interest was decreased to 4%

### Question 10

Sajan lent Rs. 7000 to two of his friends: to one at 4% and to other at 6% SI. The total amount received as interest by Sajan after 3 years is Rs. 996. Find the amount lent at 6% .

#### SOLUTION

Solution :B

Principal = Rs. 7000

Interest Received = Rs. 996

So, average rate =(996×100)3×7000=16635%

Now apply alligation:

So, the ratio in which the money was lent at 4% and 6% is 22: 13

So, the amount lent at 6% = Rs. 2600. Hence option (b)

Alternatively: the question can be solved using answer options.

### Question 11

The compound interest on a certain sum for 2 years is Rs. 756 and simple interest is Rs. 720. Now, If the sum is invested such that the SI is Rs. 1296 and the number of years is equal to the rate percentage per year, then find the rate per cent:

#### SOLUTION

Solution :C

It is given that simple interest for 2 years = Rs. 720

So, for 1 year = Rs. 7202 = Rs. 360 (because it is SI)

Compound Interest for 2 year = Rs. 756

Difference in the interest = Rs. 36 (756-720)

This difference is the interest on the interest of the first year.

So rate of interest =(36360)×100=10%

Now the interest for 1st Year = Rs. 360 at 10% SI

So, principal for 1st year = Rs. 3600

We have, SI=(P×R×T)100

3600×k2100=1296

k = 6

### Question 12

Of the two machines purchased, the first one’s share of cost is 37.5%. Depreciation is to be calculated annually at the rates of 20% and 15% on the first and second machines respectively, of the value of each machine at the beginning of the year. At the end of two years, the total value of depreciation is what percent of the total cost?

#### SOLUTION

Solution :C

Let the cost of two machines be Rs. 800

So, cost of 1st machine = 37.5% of Rs. 800 = Rs. 300

Cost of 2nd machine = 62.5% of Rs. 800 = Rs. 500

Total cost of 2 machines at the end of 2 yrs = 553.25

Total depreciation =246.75=(246.75800)×100=30.84%

Hence option (c)

### Question 13

When price of wheat increased by 44%, a family reduced its consumption in such a way that the expenditure on wheat is only 20% more than before. If 60 kg of wheat was the previous consumption. The decrease in the consumption is

#### SOLUTION

Solution :

Method 1:Quantity×Rate=Price

Given : Q1= 60

R1 =r

P1= p

Q2=q

R2=1.44r

P2=1.2p

Hence Q=50

So the difference in price is 50.

Method 2:Let the price per kg of wheat = Rs 10, hence total expenditure initially = Rs 600

The expenditure increases by 20% = Rs 720

Also, the price increases by 44% , hence price per kg = 14.4

Thus, new consumption =72014.4 = 50 kgs. the diffrence in consumption is 60-50= 10 kg. Option (a)

### Question 14

A particular shop sells goods at cost. A person buys goods worth Rs.60 from the shop and gives the shopkeeper a Rs.100 note. As the shopkeeper does not have change to give his customer, he exchanges this Rs.100 note for 6 Rs.10 notes and 2 Rs. 20 notes from the bank to settle accounts with the customer. The next day, the banker returns the Rs.100 note claiming it to be a counterfeit and takes two Rs.50 notes from the shopkeeper. If the note was actually a counterfeit, what is the total loss to the shopkeeper?

#### SOLUTION

Solution :C

The shopkeeper has lost Rs.100 only. The entire transaction can also be viewed as the shopkeeper getting a counterfeit Rs.100 note against which he has given goods worth Rs.60 and change worth Rs.40.

### Question 15

A 10% solution is one in which 10 gm of solute is dissolved in 100 gm of water. On heating the solution water evaporates. So, 1 kg of the 10% solution is reduced to 0.4 kg. What is its concentration now?

#### SOLUTION

Solution :D

In a 10% solution, in 110 gm of solution there is 10gm of solute.

So, 1 kg of solution has 1×10110=111 kg of solute.

After heating, amount of solution = 0.4 kg

So, amount of water =25−111=1755 kg

Solution percentage =(111)×(5517)×100=29.4%