Quantitative Aptitude Questions for IBPS PO Mains Exam Set – 35

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1. 24 men complete a work in 15 days and 15 women do the same work in 28 days. 18 men started the work, and worked for 10 days. After this they are replaced by 21 women. In how many days now the remaining work will be completed?
   A. 10 days
   B. 12 days
   C. 18 days
   D. 20 days
   
   Answer

   A. 10 days
   Explanation:
   24 m = 15 days. So 1 m = 24*15 days
   18 m = 24*15/18 = 20 days …. (1)
   15 w = 28 days
   So 1 w = 15*28 days
   21 w = 15*28/21 = 20 days …. (2)
   18 m worked for 10 days and their 1 day work is 1/20 by equation (1)
   Let 21 women work for ‘x’ days and their 1 day work is 1/20 by equation (2)
   So, (1/20) * 10 + (1/20) * x = 1
   Solve, x = 10 days
   
   
   
2. A can contains 3 liquids – A, B, and C in the ratio 3 : 4 : 5 respectively. 12 litres of the mixture is taken out and 8 litres of A is poured in the can. After this the ratio of A to C is 7 : 5. Find the original quantity of the mixture?
   A. 48 litres
   B. 36 litres
   C. 39 litres
   D. 60 litres
   
   Answer

   B. 36 litres
   Explanation:
   Let A = 3x, B = 4x, C = 5x
   When 12 litres taken out, left out quantities are
   A = 3x – 3/(3+4+5) * 12 = 3x – 3
   B = 4x – 4/(3+4+5) * 12 = 4x – 4
   C = 5x – 5/(3+4+5) * 12 = 5x – 5
   Now after 8 litres of A poured in, A becomes (3x – 3) + 8 = 3x + 5
   Now given, A : C = 7 : 5. So
   3x+5 / 5x-5 = 7/5
   Solve, x = 3
   Total initial quantity = 3x + 4x + 5x = 12x = 12*3 = 36
   
   
   
3. The cost price of an article is Rs 400. Some part of it is sold at 36% profit and remaining part at 24% loss. In this whole transaction, there is overall loss of 12%. Find the price of the article sold at 24% loss.
   A. Rs 80
   B. Rs 100
   C. Rs 220
   D. Rs 320
   
   Answer

   D. 320
   Explanation:
   One part at 36% profit and other at 24% loss
   By the method of allegation:
   36                                                         -24                          negative sign for loss
   .                                     -12
   -12 – (-24) = 12                              36 – (-12) = 48
   12 : 48 = 1 : 4
   So part at 24% loss is 4/(1+4) * 400 = 320
   
   
   
4. A sum of money is invested in two different schemes for 2 years. Half at 6% per annum compound interest and the remaining at 4% simple interest. The difference in the interest received by both the schemes is Rs 218. Find the principal amount.
   A. Rs 6,000
   B. Rs 8,000
   C. Rs 10,000
   D. Rs 12,000
   
   Answer

   C. Rs 10,000
   Explanation:
   Let total sum is 2P
   Half at CI, so CI = P [1+6/100]² – P
   Half at SI, so SI = P*4*2/100
   So, P [1+6/100]² – P – P*4*2/100 = 218
   Solve, P = 5000
   So total amount = 2*5000 = 10,000
   
   
   
5. There are 3 red balls, 4 white and 6 yellow. Two balls are drawn at random. What is the probability that one is red and the other is white or yellow?
   A. 5/19
   B. 2/17
   C. 5/16
   D. 5/13
   
   Answer

   D. 5/13
   Explanation:
   There are 3 red balls, 10 white and yellow and total 13 balls
   2 balls are drawn, so
   Required probability = ³C₁ * ¹⁰C₁ / ¹³C₂
   = 3*10/(13*12/2) = 5/13
   
   
6. Two die are rolled simultaneously. What is the probability that the sum of numbers on the two die is a multiple of 4?
   A. 1/4
   B. 1/2
   C. 3/4
   D. 1/3
   
   Answer

   A. 1/4
   Explanation:
   The maximum sum can be (6+6) = 12
   Up to 12, multiples of 4 are- 4, 8,12
   The pairs with sum 4 are – {(1,3)(3,1)(2,2)(2,6)(6,2)(3,5)(5,3)(4,4)(6,6)}
   These are 9 pairs. There are total 36 pairs.
   So required prob = 9/36 = 1/4
   
   
7. A and B complete a work in 12 days. B and C can complete a work in 15 days. A and C complete same work in 18 days. What is the respective ratio of the number of days taken by A to complete the same task alone to that of B completing the same task alone?
   A. 3 : 7
   B. 2 : 15
   C. 17 : 13
   D. 13 : 5
   
   Answer

   C. 17 : 13
   1/a + 1/b = 1/12
   1/b + 1/c = 1/15
   1/c + 1/a = 1/18
   Add equation 1 and 3 and subtract 2
   1/a + 1/b + 1/c + 1/a – (1/b + 1/c) = 1/12 + 1/18 – 1/15
   2/a = 13/180
   A = 360/13
   Add equation 1 and 2 and subtract 3
   1/a + 1/b + 1/b + 1/c – (1/c + 1/a) = 1/12 + 1/15 – 1/18
   2/b = 17/180
   C = 360/17
   A : C = 360/13 : 360/17 = 17: 13
   
   
   
8. There are 3 pipes. A fills the cistern in 3 hrs, B fills the cistern in 4 hrs, C empties the cistern in 1 hr. A, B, and C are opened at 3 PM, 4 PM and 5 PM respectively. At what time the cistern will get empty?
   A. 7:00 PM
   B. 7:12 PM
   C. 7:24 PM
   D. 7:36 PM
   
   Answer

   B. 7:12 PM
   Explanation:
   From 3-4 PM, cistern filled by A is 1/3
   From 4-5 PM, cistern filled by A + B is 1/3 +1/4 = 7/12
   Total filled up to 5 PM = 1/3 + 7/12 = 11/12
   Now after 5 PM, when C is also opened, work done in 1 hr = 1/3 + 1/4 – 1 = – 5/12
   This means that after 5 PM, 5/12 cistern is emptied in 1 hr.
   So 11/12 cistern will be emptied in (11/12) / (5/12) = 11/5 hrs = 2 hrs 12 min
   So total time 5 PM + 2 hrs 12 min = 7:12 PM
   
   
9. The volume and curved surface area of a cylinder is 5390 cm³ and 1540 cm². Find the total surface area of the cylinder.
   A. 1848 cm²
   B. 1500 cm²
   C. 1600 cm²
   D. 1765 cm²
   
   Answer

   A. 1848 cm²
   Explanation:
   Vo. of cylinder is = ᴨr²h = 5390
   Curved surface area = 2ᴨrh = 1540
   ᴨr²h / 2ᴨrh = 5390/1540
   Solve, r = 7, put r in 2ᴨrh = 1540, h= 35
   total surface area = 2ᴨrh + 2ᴨ r²
   put the values.
   
   
   
10. Surface area of a cube is 600 cm². Find the length of the diagonal of the cube.
    A. 35 cm
    B. 50 cm
    C. 20√3 cm
    D. 10√3 cm
    
    Answer

    D. 10 √3
    Explanation:
    6a² = 600
    a = 10
    Diagonal = √3 a = 10 √3