1. The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder. What is the smaller number ?

A. 240 B. 270 C. 295 D. 360

Answer - Option B
Explanation - Let the smaller number be x. Then larger number = (x + 1365).
i.e, x + 1365 = 6x + 15
[latex]\Rightarrow[/latex] 5x = 1350
[latex]\Rightarrow[/latex] x = 270
Smaller number = 270.
2. 72519 x 9999 = ?

A. 725117481 B. 674217481 C. 685126481 D. 696217481

Answer - Option A
Explanation - 72519 x 9999 = 72519 x (10000 - 1)
= 72519 x 10000 - 72519 x 1
= 725190000 - 72519
= 725117481
3. If the number 517*324 is completely divisible by 3, then the smallest whole number in the place of * will be:

A. 0 B. 1 C. 2 D. None of these

Answer - Option C
Explanation - Sum of digits = (5 + 1 + 7 + x + 3 + 2 + 4) = (22 + x), which must be divisible by 3.
x = 2.
4. In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is [latex]\frac {2}{3}[/latex] of the number of students of 8 years of age which is 48. What is the total number of students in the school?

A. 72 B. 80 C. 120 D. 100

Answer - Option D
Explanation - Let the number of students be x. Then,
Number of students above 8 years of age = (100 - 20)% of x = 80% of x.
i.e, [latex]80% of x = 48 +\frac {2}{3}[/latex]of 48
[latex]\Rightarrow \frac {80}{100} x = 80[/latex]
x = 100.
5. In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was:

A. 2700 B. 2900 C. 3000 D. 3100

Answer - Option A
Explanation - Number of valid votes = 80% of 7500 = 6000.
i.e, Valid votes polled by other candidate = 45% of 6000
= [latex](\frac {45}{100} \times 6000) = 2700[/latex]

1. What will be the compound interest on a sum of Rs. 25,000 after 3 years at the rate of 12 p.c.p.a.?

A. Rs. 9000.30 B. Rs. 9720 C. Rs. 10231.20 D. Rs. 10483.20

Answer - Option C
Explanation - Amount = Rs.[latex][25000 \times {(1 + \frac {12}{100})]}^{3}[/latex]
= Rs[latex] (25000 \times \frac {28}{25} \times \frac {28}{25} \times \frac {28}{25})[/latex]
= Rs. 35123.20
i.e, C.I = Rs.(35123.20 - 25000) = Rs. 10123.20
2. Albert invested an amount of Rs. 8000 in a fixed deposit scheme for 2 years at compound interest rate 5 p.c.p.a. How much amount will Albert get on maturity of the fixed deposit?

A. Rs. 8600 B. Rs. 8620 C. Rs. 8820 D. Rs. 8828

Answer - Option C
Explanation - Amount = Rs.[latex][8000 \times {(1 + \frac {5}{100})]}^{2}[/latex]
= Rs[latex] (8000 \times \frac {21}{20} \times \frac {21}{20})[/latex]
= Rs. 8820
3. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:

A. Rs. 650 B. Rs. 690 C. Rs. 698 D. Rs. 700

Answer - Option C
Explanation - S.I. for 1 year = Rs. (854 - 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
i.e, Principal = Rs. (815 - 117) = Rs. 698.
4. Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?

A. Rs. 6400 B. Rs. 6500 C. Rs. 7200 D. Rs. 7500

Answer - Option A
Explanation - Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x).
Then, [latex](\frac {x \times 14 \times 2}{100}) + (\frac {(13900 - x) \times 11 \times 2}{100}) = 3508[/latex]
28x - 22x = 350800 - (13900 x 22)
6x = 45000
x = 7500.
So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.
5. The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is:

A. 15 B. 16 C. 18 D. 25

Answer - Option B
Explanation - Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.
S.P. of x articles = Rs. 20.
Profit = Rs. (20 - x).
i.e, [latex](\frac {20 - x}{x} \times 100 = 25)[/latex]
[latex]\Rightarrow[/latex] 2000 - 100x = 25x
125x = 2000
x = 16.

1. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?

A. Rs. 4991 B. Rs. 5991 C. Rs. 6001 D. Rs. 6991

Answer - Option A
Explanation - Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.
i.e, Required sale = Rs. [ (6500 x 6) - 34009 ]
= Rs. (39000 - 34009)
= Rs. 4991.
2. The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?

A. 0 B. 1 C. 10 D. 19

Answer - Option D
Explanation - Average of 20 numbers = 0.
i.e, Sum of 20 numbers (0 x 20) = 0.
It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).
3. The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?

A. 76 kg B. 76.5 kg C. 85 kg D. None of these

Answer - Option C
Explanation - Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.
4. The ratio of the number of boys and girls in a college is 7 : 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio?

A. 8 : 9 B. 17 : 18 C. 21 : 22 D. None the these

Answer - Option C
Explanation - Originally, let the number of boys and girls in the college be 7x and 8x respectively.
Their increased number is (120% of 7x) and (110% of 8x).
[latex]\Rightarrow (\frac {120}{100} \times 7x) [/latex]and [latex](\frac {110}{100} \times 8x)[/latex]
[latex]\Rightarrow \frac {42x}{5}[/latex] and [latex] \frac {44x}{5} = 21 : 22[/latex]
i.e, The required ratio = [latex](\frac {42x}{5} : \frac {44x}{5})[/latex] = 21 : 22
5. Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40 : 57. What is Sumit's salary?

A. Rs. 17,000 B. Rs. 20,000 C. Rs. 25,500 D. Rs. 38,000

Answer - Option D
Explanation - Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then, [latex]\frac {2x + 4000}{3x + 4000} = \frac {40}{57}[/latex]
[latex]\Rightarrow[/latex] 57(2x + 4000) = 40(3x + 4000)
[latex]\Rightarrow[/latex] 6x = 68,000
[latex]\Rightarrow[/latex] 3x = 34,000
Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000.