Quant Test for IBPS Clerk 2018 Main Exam Set – 27

Directions(1-5): What will come in place of question mark “?” in the following questions.

1. 14(3/5) + 8(2/5) – 5(4/5) = (?)^1/2 + 6(7/15)
   90
   98
   100
   88
   121
   Option E
   14(3/5) + 8(2/5) – 5(4/5) = (?)^1/2 + 6(7/15)
   => (?)^1/2 + 97/15 = 73/5 + 26/3 – 29/5
   => (?)^1/2 = 44/5 + 26/3 – 97/15
   => ? = 121

    

2. 17.98 * 7(1/7)% of 363.98 + 39.98% of 519.98 = ?^2
   15
   20
   18
   26
   10
   Option D
   17.98 * 7(1/7)% of 363.98 + 39.98% of 519.98 = ?^2
   => ?^2 = 18*1/14*364+0.4*520
   => ?^2 = 18*26+208
   => ? = 26

    

3. 47*53 – (64*4 – (841)^1/2) = ? * 250% of 3.2
   277
   283
   290
   250
   281
   Option B
   47*53 – (64*4 – (841)^1/2) = ? * 250% of 3.2
   => ? * 8 = 2500 -9 – (256 – 29)
   => ? = 283

    

4. (5475.98)^1/2 * 39.98 + 37.5% *1399.87 = ?^2 + 1.99^2
   59
   44
   48
   50
   35
   Option A
   (5475.98)^1/2 * 39.98 + 37.5% *1399.87 = ?^2 + 1.99^2
   => ?^2 = (5476)^1/2 *40 + 3/8*1400 – 4
   => ?^2 = 74*40 + 525 – 4
   => ? = 59

    

5. (16.99% of 799.7)*24.98 = ?^2 – 33.33% of 962.88
   61
   55
   40
   66
   70
   Option A
   (16.99% of 799.7)*24.98 = ?^2 – 33.33% of 962.88
   => ?^2 = (17% of 800)*25 +963/3
   => ?^2 = 136*25 + 321
   => ? = 61

    

6. Two containers A and B of equal capacities contain milk and water in the ratio of 3:2 and 2:1 resp. If (x+15)% of milk solution is taken out from container A and added to container B becomes 177:93 resp. Find the value of x.
   8
   5
   6
   4
   9
   Option B
   Let the quantity of the mixture in container A and B be y each.
   After transferring milk and water from A to B one by one [2y/3+3y(x+15)%/5]/[y/3+2y(x+15)%/5] = 177/93
   => x = 5

    

7. Shyam gives away 320 toffees to five children, Rani,Pinki,Raju, Toni and Emma. The ratio of toffees received by Toni, Emma and Rani is 3: (x+1): 6 resp. and the ratio of toffees received by Rani , Pinki and Raju is 3:x:5 resp. Find the value of x, if Rani and Toni together received 90 toffees.
   12
   4
   10
   8
   15
   Option B
   Ratio of toffees received by them = 3: (x+1) : 6 : 2x : 10
   Now,(3+6)/(3+x+1+6+2x+10) = 1/32
   => x = 4

    

8. The age of Batsman, Spiderman and Antman is (x+4) years, (x-y+10) years and (y+4) resp. The ratio of age of Batsman after 4 years and age of Spiderman 6 years ago is 3:1 resp. and age of Spiderman after 3 years will be equal to the age of Antman 3 years ago. Find the value of (x+y).
   48
   62
   51
   40
   70
   Option A
   [(x+4)+4]/[(x-y+10)-6] = 3/1 => 2x – 3y + 4 = 0 —–(1)
   Also, (x-y+10)+3 = (y+4)-3
   => 2x – 4y + 24 = 0 —–(2)
   Subtracting (1) from (2), we get
   x = 28
   y = 20
   (x+y) = 48

    

9. A and B can do a piece of work in (x-2) days and (x+2) days resp. while working alone. If A and B together can complete the whole work in 24/5 days. Find the sum of time taken by A and B to complete the work individually.
   50 days
   30 days
   20 days
   10 days
   40 days
   Option C
   [(x-2)(x+2)]/[(x-2)+(x+2)] = 24/5
   => 5x^2 – 48x – 20 = 0
   => 5x^2 – 50x + 2x – 20 = 0 => 5x(x-10) + 2(x-10) = 0 => x = 10
   Time taken by A to complete the work alone = 10-2 = 8 days
   Time taken by B to complete the work alone = 10+2= 12 days
   Sum of the time taken by both to complete the work = 8+12 = 20 days

    

10. A packet of balloons contains 18 balloons of three colours: red, yellow and green. The probability of choosing either a red or yellow balloon is 2/3. If 3 balloons are choosen at random, find the probability that at most 2 balloons are green.
    193/200
    187/211
    193/201
    199/204
    191/200
    Option D
    Number of red and yellow balloons = 2/3*18 = 12
    Number of green balloons = 6
    Now, Required probability = [12C3+12C1*6C2+12C2*6C1]/18C3
    = 199/204