Aptitude

Showing 31 - 40 of 116 questions

31

The remainder when 1010+ 10100+101000+ . . . . . . + 101000000000 is divided by 7 is

a

0

b

1

c

2

d

5

Number of terms in the series = 10.
(We can get it easily by pointing the number of zeros in power of terms.
In 1st term number of zero is 1, 2nd term 2, and 3rd term 3 and so on.)

1010/7, Written as,
(7+3)(4*2+2)/7
The remainder will depend on,
32/7
So, remainder will be 2.
10100/7, Can be written as,
3100/7.
Or, 325*4/7
Or, 34/7.
Remainder = 1
101000/7, Remainder = 2
1010000/7, Remainder = 1

So, we get alternate 2 and 1 as remainder, five times each.
So, required remainder is given by

(2+1+2+1+2+1+2+1+2+1)/7
= 15/7

Remainder: 1

32

The number of oranges in three basket are in the ratio 3 : 4 : 5. In which ratio the no. of oranges in first two basket must be increased so that the new ratio becomes 5 : 4 : 3 ?

a

3:4

b

2:1

c

2:3

d

1:3

Let,
B1 : B2 : B3 = 3x : 4x : 5x
Again,
B1 : B2 : B3 = 5y : 4y : 3y
Number of oranges remain constant in third basket as increase in oranges takes place only in first two baskets.
Hence, 5x = 3y
x=(3y)/5
and,
3x : 4x : 5x
→ (9y) /5 : (12y) /5 : (15y) /5 = 9y : 12y : 15y
And,
5y : 4y : 3y → 25y : 20y : 15y

Therefore, Increment in first basket = 16.
Increment in second basket = 8.
Thus, Required ratio = 16 /8 = 2 : 1

33

The cost of setting up a magazine is Rs. 2800. The cost of paper and ink etc is Rs. 80 per 100 copies and printing cost is Rs. 160 per 100 copies. In last month 2000 copies were printed but only 1500 copies could be sold at Rs. 5 each. Total 25% profit on the sale price was realized. There is one more resource of income from magazine which is advertising. What sum of money obtained from the advertising in magazine?

a

Rs.1750

b

Rs.2350

c

Rs.1150

d

Rs.1975

Set up cost = Rs. 2800
Paper etc = Rs. 1600
Printing cost = Rs. 3200
Total cost = Rs. 7600
Total sale price = 1500*5 = 7500
Let amount obtained from advertising be x then,
(7500+x)-7600 = 25 % of 7500
x = 1975.

34

Today is Monday. After 63 days, it will be :

a

Tuesday

b

Monday

c

Sunday

d

Saturday

Each day of the week is repeated after 7 days.

63/7=9, Remainder:0

So, after 63 days, it will be Monday.

35

Avinash covers a distance of 8km in 50 minutes. If he covers 3km distance in 2/5th of time then, What speed should he maintain to cover the remaining distance in the remaining time?

a

13 Kmph

b

14 Kmph

c

15 Kmph

d

11 Kmph

Total distance = 8 Km

Total time = 50 minutes

Time taken to cover the distence of 3 Km = 50**(2/5) = 20 min = 1/3 hours

Remaining distance = 8 - 3 = 5 Km

Required speed = 5/((1/3)) = 15 Kmph

36

Fresh fruit contains 68% water and dry fruit contains 20% water. How much dry fruit can be obtained from 100 kg of fresh fruits ?

a

20

b

30

c

40

d

50

The fruit content in both the fresh fruit and dry fruit is the same.

Given, fresh fruit has 68% water. So remaining 32% is fruit content. Weight of fresh fruits is 100kg

Dry fruit has 20% water. So remaining 80% is fruit content. Let weight of dry fruit be y kg.

Fruit % in freshfruit = Fruit% in dryfruit

$\inline \fn_cm \therefore$(32/100) ** 100 = (80/100 ) ** y

We get, y = 40 kg

37

In a two-digit number, the digit in the unit's place is more than twice the digit in ten's place by 1. If the digits in the unit's place and the ten's place are interchanged, difference between the newly formed number and the original number is less than the original number by 1. What is the original number ?

a

35

b

36

c

37

d

39

Let the ten's digit be x. Then, unit's digit = 2x + 1.

[10x + (2x + 1)] - [{10 (2x + 1) + x} - {10x + (2x + 1)}] = 1

<=> (12x + 1) - (9x + 9) = 1 <=> 3x = 9, x =  3.

So, ten's digit = 3 and unit's digit = 7. Hence, original number = 37.

38

Find the odd man out?

396, 462, 572, 427, 671, 264.

a

671

b

462

c

427

d

264

In each number except 427, the middle digit is the sum of other two.

39

Three numbers are in the ratio 4 : 5 : 6 and their average is 25. The largest number is :

a

30

b

40

c

50

d

60

Let the numbers be 4x, 5x and 6x.

Then, (4x + 5x + 6x ) / 3 = 25

=> 5x = 25

=> x = 5.

Largest number  6x = 30.

40

The diagonal of a rectangle is $\inline \sqrt{41}$ cm and its area is 20 sq. cm. The

perimeter of the rectangle must be:

a

18

b

28

c

38

d

48

sqrt(l^2+b^2) = sqrt(41) (or) l^2+b^2 = 41

Also, lb = 20

(l+b)^2 = l^2 + b^2 + 2lb

= 41 + 40 = 81

(l + b) = 9

Perimeter = 2(l + b) = 18 cm