# Sample Aptitude Questions of Nokia

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1. Simple interest on Rs. 1200 @ 13 p.c.p.a. for 'X' years is Rs. 624/-. What is the amount on Rs. 'X+1000' at the same rate of interest for 3 years?
1. Rs. 1872/-
2. Rs. 1384/-
3. Rs. 936/-
4. Other than those given as options
5. Rs. 1404/-

624 = (1200×13×x)/100 => x = 4
Now, P = x + 1000 = 4 + 1000 = 1004
I = (1004×13×3)/100 = 392
=> amount = 1004 + 392 = 1396/-
2. A circle and rectangle have the same perimeter. The sides of the rectangle are 18 cm and 26 cm. What will be the area of the circle? (in cm2)
1. 88
2. 1250
3. 154
4. 128
5. Other than those given as options

Perimeter of rectangle = perimeter of circle = 2 (18 + 26) = 88
2Πr = 88 => r = 14cm
Area of circle = 22/7 ×14 × 14 = 616 cm2
1. 101 + 1001 + 2003 + 30005 + 9056 =?
1. 42616
2. 42166
3. 41266
4. 42156
5. 42661

=>101 + 1001 + 2003 + 30005 + 9056 = 42166.
2. 5/6th of 348 -1/8th of 232 = ?
1. 267
2. 258
3. 257
4. 261
5. 263

=>5/6th of 348 -1/8th of 232 = x
=> x = 290 – 29 = 261
3. 3060 -2460 = ? × 30
1. 30
2. 50
3. 20
4. 60
5. 43

=>3060 -2460 = x * 30
=> 600 = 30x => x = 20
4. The product of the digits of a two-digit number is twice as large as the sum of its digits. If we subtract 27 from the required number, we get a number consisting of the same digits written in the reverse order. Find the number.
1. 36
2. 27
3. 63
4. 46
5. None of these

Go by options. 3rd option is the answer because 63 Þ product of digits = 6*3 = 18. Sum of digits = 6 + 3 = 9.
Hence product of digits is twice as the sum of the digits. Also 63 – 27 = 36. So digits are reversed.
5. The product of the digits of a two-digit number is one-third of that number. If we add 18 to the required number we get a number consisting of the same digits written in reverse order. Find the number.
1. 42
2. 24
3. 72
4. 27
5. None of these

Go by options 2nd option correct because 24 = 3*2*4 . Number is thrice the product of its digits
24+18 = 42. Hense digits are reversed.
6. Jar A has 60 litres of mixture of milk and water in the respective ratio of 2: 1. Jar B which had 40 litres of mixture of milk and water was emptied into Jar A, as a result in Jar A, the respective ratio of milk and water becomes 13 : 7. What was the quantity of water in Jar B?
1. 8 litres
2. 15 litres
3. 22 litres
4. 7 litres
5. 1 litres
Jar A has 60 Liters.
Radio between milk and water 2:1;
Quantityof milk in Jar A=2/3*60 = 40;
Quantityof waterin Jar A=1/3*60 = 20;
40 Liters of Mixture B having Milk and Water is empty;
Therefore ,total Mixture = 60 + 40 =100;
The respective Ratio of Milk and Water is 13:7;
Quantity of Milk in Jar A = 13 /20 *100 = 65;
Quantity of Waterin Jar A = 7/20 *100 = 35;
Quantity of Waterin Jar B = 35 -20 = 15 liters.
7. The sum of a series of 5 consecutive odd numbers is 195. The second lowest number of this series is one less than the second highest number of another series of 5 consecutive even numbers. What is 40% of the second lowest number of the series of consecutive even numbers?
1. 16.8
2. 14.8
3. 19.4
4. 17.6
5. 13.6

The sum of the series of 5 Consecutive odd numbers is 195;
Let the Series of Consecutive odd numbers is X, X+2, X+4, X+6, X+8 ;
X+X+2+X+4+X+6+X+8 = 195 ;
5X+20 = 195;
X=35.
Series of consecutive odd numbers is 35, 37 , 39 , 41 ,43 ;
According to the question ,the second lowest number of this series is less than the ;
Second highest number of another series of 5 consecutive numbers;
Second lowest number is 37;
Second highest number of another series of 5 consecutive numbers = 37 +1 = 38 ;
Therefore , another series of 5 consecutive no 32 , 34 ,36 ,38 , 40 ;
40% of the lowest of the series of consecutive numbers = 34 *40 / 100 = 13.6.
8. The sum of the dimensions of a room (i.e. length, breadth and height) is 18 metres and its length, breadth and height are in the ratio of 3 : 2 : 1 respectively. If the room is to be painted at the rate of Rs. 15 per m2, what would be the total cost incurred on painting only the four walls of the room (in Rs.)?
1. 3250
2. 2445
3. 1350
4. 2210
5. 2940

Ratio of length : breadth : height = 3 : 2 : 1
Sum of dimensions of room = 18
Length = 3/6×18 = 9
Breath = 2/6×18 = 6
Height = 1/6×18 = 3
Area of four walls = 2h×(l+b) = 2×3(9+6) = 90
Total cost of painting four walls = 90×15 = 1350
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