# Free Basic Arithmetic - 01 Practice Test - CAT

### Question 1

To build a house at a faster pace, the contractor increases the wages/ hour of each worker by 20% and the working hours by 25% . By how much percent has the daily earnings of each labourer increased?

#### SOLUTION

Solution :D

Solve using assumption

Lets say the number of hours / day initially = 10 and the wages = Rs 100 for 10 hours

Wages / hour = 10010 = 10/hr

Earnings initially = 100 for 10 hours

New wages = 1.2×10 = 12 / hr

New number of hours = 1.25×10 = 12.5 hours

New earnings = 12×12.5 = 150

% change =50100×100=50% , Option (4).

2nd method:-Wages(w)=no.ofhours(h)×wages/hour(p)W2W1=h2×p2h1×p1=1.25h1×1.2p1h1×p1=1.2×1.25=1.5

F.O.M = 1.5 i.e % increase = 50%. Option (4) .

### Question 2

For construction, a certain amount of wire was being shipped, out of which 10% was stolen. After using 85% of the remainder, 472.5 m of wire was left. How much wire was initially shipped?

#### SOLUTION

Solution :D

Take a middle answer option and verify the answers.

Initial = 3500

Amount left after being stolen =0.9×3500=3150

Amount left =15% of 3150=472.5

2nd method:-Let the total length of original wire = xRemaining wire =100%−10% (Stolen wire)=90% . After using 85% of remaining wire, 15% of remaining wire was left.

15% of 90% 0f x=472.5

0.135 of x=472.5⇒x=3500.

### Question 3

If the side of a square is increased by 25% , then its area is increased by how many percent?

#### SOLUTION

Solution :C

Let the side be 10 cm. Then the area will be 100cm2

New side =125% of 10=12.5cm;

Area=(12.5)2=156.25

Percentage increase in area =56.25%.

Alternatively:If x is the percentage increase in the side of a square, then increase in area is given by

x+x+x×x100=2x+x2100=25+25+25×25100=56.25%

### Question 4

In a parliamentary debate, a bill was being passed. 600 votes were cast initially but after some discussion, the opposition to the bill went up by 150%. The bill was then rejected by a majority 2 times as great as that by which it was passed previously. How many people rejected the bill initially?

#### SOLUTION

Solution :D

There are totally 600 voters, so after the number of people who initially rejected increases by 150%, it must still be below 600. That is true only for option (d).

Verifying the answer chosen:

Initially,

Against = 200, for = 400. Hence, majority = 400-200 = 200

After discussions,

Against =200+1.5×200=500 , for = 600-500 = 100. Hence, majority = 500-100 = 400, which is twice the initial majority. Thus, this is correct answer.

Alternatively:There are totally 600 voters. Let number of peole, who were in Against = A, and who were in favour = F. Hence, majority = F - A

F + A = 600

After discussions,

The opposition to the bill went 150% . Aganist = 2.5 A, Favour =600−2.5A

According to given question,

2[600−2.5A]=[600−2A]⇒A=200 and F=400.

### Question 5

Fresco has doubled its turnover in 2000 from 5 crores in 1999, tripled its turnover in 2001 and grew by 50% in 2002.The turnover at the end of 2002 is

#### SOLUTION

Solution :1999 = 5 crores

2000=5×2=10 crores

2001=10×3=30 crores

2002=30×1.5=45 crores

### Question 6

In a college 50% students are boys out of which 40% have got 80% and above marks. Overall, 45% of all the students have got 80% and above marks. Find the percentage of girls who have got less than 80% marks.

#### SOLUTION

Solution :C

Suppose 100 students are in the college: 50 boys and 50 girls. Now 40% of the boys i.e. 20 boys have got 80% and above marks. And Total 45 students have got 80% and above marks. So, number of girls who have got 80% and above = 25. Total number of girls = 50. Hence 50% of the girls have got less than 80% marks.

### Question 7

20% of a larger number is 2.3 less than 30% of a smaller number. The larger number also exceeds the smaller number by 10. What is the value of the larger number?

#### SOLUTION

Solution :E

Let the larger number and smaller number is L and S respectively.

0.2L=0.3S−2.3 & 2L−3S=−23

L−S=10 & 2L−2S=20

Solving the above two equations we get L = 53. Hence option (e)

### Question 8

Rita scored 25% less than Nita. How much percent more did Nita score than Rita?

#### SOLUTION

Solution :C

Approach 1:

%Nita scored more than Gita =25100−25×100%=33.33%

Approach 2:

Rita’s score is lesser by 25% (14). Nita’s score will be greater by 1x−1%=(13)=33.33%

### Question 9

Laxman had 100 cows at the beginning of the year 1992 and the number of cows each year increases by 10%. At the end of each year he doubles the present no. of cows by buying extra cows. What is the number of cows he has at the beginning of 1994?

#### SOLUTION

Solution :E

Solution:

At the beginning of the year 1992, He was having 100 cows.

At the beginning of the year 1993, Number of cows =2(1.1×100)=220.

At the beginning of the year 1994, Number of cows =2(1.1×220)=484.

### Question 10

Anil and Binod started a business by investing Rs.35,000 and Rs.13,000 respectively. At the end of every month, Anil withdraws certain amount from his investment and Binod invests the same amount as Anil has withdrawn. At the end of the year they share the profits in the ratio 1:1 . Find the amount invested by Binod every month.

#### SOLUTION

Solution :B

Let x be the Amount that Anil withdraws and Binod invests at the end of every month.

Then, 35,000−11x=13,000.

11x=22,000 & x=2000.

Alternatively:-1stYear2ndYear3rdYear.............12thYearAnil35,000(35,000−x)(35,000−2x).............(35,000−11x)Binod13,000(13,000−x)(13,000−2x).............(13,000−11x)

35,000×12−66x=13,000×12+66x

22,000×12=132x

x = 2000.

### Question 11

A started a business with Rs. 52,000 and after 4 months B joined him with Rs.39,000. At the end of the year, out of the total profits B received total Rs. 20,000 including 25% of the profits as commission for managing the business. What amount did A receive?

#### SOLUTION

Solution :C

Profit's share of A and B =(52,000×12)(39,000×8)=2:1

Let the profit be Rs. x, then B receives 25% as commission for managing business, the remaining 75% of the total profit x is shared between A and B in the ratio 2:1. Hence B will get 13rd part of this in addition to his commission. Hence his total earning

=0.25x+13×0.75x

=0.5x=20,000

x=40,000

So,the remaining profit goes to A,hence the profit of A is Rs.20,000.

### Question 12

Pankaj takes a loan of Rs. 10000 at 10% per annum compounded annually which is to be repaid in two equal annual instalments: one at the end of the first year and second at the end of the second year. The value of each instalment is:

#### SOLUTION

Solution :A

Let the value of each installment is X, Total amount is P

P=X(K+K2+K3+...................),Where K=100100+r Here only two years and r is 10% then

P=X(K+K2)

10000=x{1011+(1011)2}

X = 5761.9

Alternatively10,000(1.1)2=x[1+(1.1)]

x = 5761.9

### Question 13

Deepak invested Rs. 50000 in a bond which was giving him 10% interest per annum at compound interest. Find the money received by Deepak after 3 years if he has to pay 20% tax on the compound interest every year.

#### SOLUTION

Solution :E

Deepak was getting 10% interest, 20% of the interest i.e. 20% of 10% (=2%) was the tax which he had to pay, so the net interest rate (after tax) =10−2=8%

So, amount received by him after 3 years =50000(1+(8100))3=Rs.62985.6

Hence option (5)

2nd method:-By, pascal triangles for n=3 & r=8%. Option (5).

### Question 14

A certain amount of money has been invested to earn a simple interest of 10% per annum. The interest earned every year is again invested at the same rate. After how many years will the amount become the square of itself?

#### SOLUTION

Solution :E

Unless we know the amount we cannot derive the answer.

Hence option (e)

### Question 15

On a certain sum, the difference between the compound interest and the simple interest for the second year is Rs.4500 and the same for the third year is Rs.14175. What is the sum?

#### SOLUTION

Solution :A

The difference in the SI and CI for the second year =Pr21002=4500....................(i)

The difference in the SI and CI for the third year =Pr21002(3+r100)=14175...........(ii)

From the equation (i) and (ii).

4500×(3+r100)=14175

r100=(141754500−3)

r=67545. Now ,from equation (i)

P=4500r2×1002⇒P=4500675×675×1002×45×45⇒P=2,00,000.