Quantitative Aptitude Practice Question and Answer
8Q: A duster is of dimension 5x5x7 all in cms, is sinked into a cylindrical vessel of radius 5 cm and height 10 cm. After insertion the level of the water is inccresed by 2.14 cm. Then by what percentage the part of the duster is above the layer ? 1600 05b5cc705e4d2b4197774f216
5b5cc705e4d2b4197774f216- 196 %false
- 24 %true
- 397 %false
- 43 %false
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Answer : 2. "4 %"
Explanation :
Answer: B) 4 % Explanation: Imension of the duster is assumed as 5x5x7 cm3 Volume of the duster below the water layer= volume of water increased= πxrxrxh = 3.14×52×2.14 ≈ 168 cm3 Total volume of the duster = 5×5×7 = 175 cm3 Percentage of the duster part below the layer = 168×100/175 = 96% Percentage of the duster part above the layer = 4%
Q: The area of a square is equal to the area of a rectangle. The length of the rectangle is 5 cm more than a side of the square and its breadth is 3 cm less than the side of the square. What is the perimeter of the rectangle ? 1599 05b5cc6c0e4d2b4197774db0f
5b5cc6c0e4d2b4197774db0f- 115 cmfalse
- 218 cmfalse
- 334 cmtrue
- 426 cmfalse
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Answer : 3. "34 cm"
Explanation :
Answer: C) 34 cm Explanation: Let the side of the square be 's' cm length of rectangle = (s+5) cm breadth of rectangle = (s-3)cm (s+5) (s-3) = s2 - 5s - 3s - 15 = s2 2s = 15 Perimeter of rectangle = 2(L+B) = 2(s+5 + s–3) = 2(2s + 2) = 2(15 + 2) = 34 cm
Q: What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ? 1599 05b5cc723e4d2b4197774f65e
5b5cc723e4d2b4197774f65e- 1844false
- 2814true
- 3840false
- 4820false
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Answer : 2. "814"
Explanation :
Answer: B) 814 Explanation: Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.Area of each tile = (41 x 41) sq.cm Required number of tiles =1517 x 902/(41 x 41) = 814.
Q: The length of a rope in meter by which a cow must be tethered in order that she may be able to graze an area of 2826 sq.m is ? 1597 05b5cc731e4d2b4197774f823
5b5cc731e4d2b4197774f823- 115 mtsfalse
- 230 mtstrue
- 324 mtsfalse
- 436 mtsfalse
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Answer : 2. "30 mts"
Explanation :
Answer: B) 30 mts Explanation: The cow can graze an area of 2826 sq.m i.e it is able to cover area of circle of 2826 sq.m Therefore, πr2 =2826r2 =(2826×7/22) = 899.18 = 900r = 30 m. Therefore, the radius of the circle implies the length of the rope = 30 mts.
Q: If in the following set of numbers, the first and the third digits are interchanged in each number, which number will be second from the end if arranged in ascending order after interchanging the digits ? 585 546 514 404 206 369 1594 05b5cc73de4d2b4197774f92f
5b5cc73de4d2b4197774f92f- 1404false
- 2585false
- 3415true
- 4602false
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Answer : 3. "415"
Explanation :
Answer: C) 415 Explanation: Given numbers are 585 546 514 404 206 369Then interchanging the 1st and 3rd digits of the given numbers, we get585 645 415 404 602 963Now, arranging them in ascending order963 645 602 585 415 404Then the last second number is 415.
Q: What is 15 as a decimal? 1592 05b5cc6aae4d2b4197774cf8e
5b5cc6aae4d2b4197774cf8e- 10.333false
- 20.666false
- 30.200true
- 40.452false
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Answer : 3. "0.200"
Explanation :
Answer: C) 0.200 Explanation: Here I converted the original fraction to a fraction with the denominator being 10, because decimals can be written in the tenths place. Divide numerator and denominator with '2'15 x 22 = 210 = 0.2
Q: 3/4 divided by 2 is 1587 05b5cc666e4d2b4197774bed0
5b5cc666e4d2b4197774bed0- 13/2false
- 22/3false
- 38/3false
- 43/8true
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Answer : 4. "3/8"
Explanation :
Answer: D) 3/8 Explanation: 3/4 divided by 2 = 3/4/2 = 3/4x2 = 3/8. Hence, 3/4 divided by 2 is equal to 3/8.
Q: On the occasion of New Year, each student of a class sends greeting cards to the others. If there are 21 students in the class, what is the total number of greeting cards exchanged by the students? 1587 05b5cc6b9e4d2b4197774d757
5b5cc6b9e4d2b4197774d757- 1380false
- 2420true
- 3441false
- 4400false
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Answer : 2. "420"
Explanation :
Answer: B) 420 Explanation: Given total number of students in the class = 21 So each student will have 20 greeting cards to be send or receive (21 - 1(himself)) Therefore, the total number of greeting cards exchanged by the students = 20 x 21 = 420.