Tuesday, 22 November 2016




1. 5, 8, 17, 24, 37, _

A. 48
B. 50
C. 64
D. 58
E. 60

Sol. 2^2 + 1, 3^2 – 1, 4^2 + 1, 5^2 – 1, 6^2 + 1

2. 1, 4, 11, 30, 85, _

A. 385
B. 205
C. 248
D. 125
E. 141

Sol. 3^0 + 0, 3^1 + 1, 3^2 + 2, 3^3 + 3, 3^4 + 4

3. 2, 4, 7, 12, 19, _

A. 40
B. 28
C. 26
D. 30
E. 37

Sol. Prime number addition

4. 3, 5, 13, 49, 241, _

A. 1372
B. 1441
C. 761
D. 1094
E. None of these

Sol. 3*2 – 1, 5*3 – 2, 13*4 – 3, 49*5 – 4, 241*6 – 5

5. 21, 10, 9, 12, 22, _

A. 38.5
B. 50.5
C. 52.5
D. 62.5
E. 78.5

Sol. 21*0.5 – 0.5, 10*1 – 1, 9*1.5 – 1.5, 12*2 – 2

6. Mixture A contains 75% milk and mixture B contains 90% milk. A milkman takes some mixture from mixture A and mixes it with four times the amount of mixture from mixture B. What is the percentage of milk in the new mixture ?

A. 90%
B. 75%
C. 87%
D. 84%
E. 78%

Sol. Let x from A, 4x form B
So milk from A = (75/100)*x, milk from B = (90/100)*4x
So milk in new mixture is (75x/100) + (360x/100) = 4.35x
Total mixture in third is x+4x = 5x
So % of milk is (4.35x/5x)*100

7. 5 men started the work and worked for 4 days. Now they are replaced by 12 women where 1 man can do as much work as 2 women do in 1 day. If the works gets completed in a total of 10 days, in how many days 7 women can complete the entire work ?

A. 20
B. 12
C. 16
D. 22
E. 14

Sol. Given 1 m = 2w, so 12 women = 6 men
Let 1 man do work in x days, so 1 man’s 1 day work = 1/x, so 5 men 1 day’s work = (1/x)*5
And since they did work for 4 days, so there work becomes (5/x)* 4 = 20/x
Now total work completed in 10 days, this mean that 12 women or 6 men did in (10-4) = 6 days
So similarly as above, 1 man’s 1 day work = 1/x, so 6 men 1 day’s work = (1/x)*6
And since they did work for 6 days, so there work becomes (6/x)* 6 = 36/x
So (20/x) + (36/x) = 1
Solve, x = 56
1 man do work in 56 days, so 1 woman in 2*56 days, and 7 women in (2*56)/7

8. The incomes of 2 persons A and B is in the ratio 4 : 5. Their expenditures are in the ratio 2 : 3 respectively. If both save Rs 50,000, then what is B’s income ?

A. 95000
B. 100000
C. 125000
D. 145000
E. 130000

Sol. Incomes 4x and 5x
Expenditures 2y and 3y
So 4x – 2y = 50000
And 5x – 3y = 50000
Solve both equations, x = 25,000
So B’s income = 5*25000

9. The radius of sphere is 16 cm and cost of painting its surface Rs 50 per square cm. Suppose the radius of sphere and its cost are increased by 5% and 10% respectively. What will be the percentage increase in the total cost of painting per square cm ?

A. 22.3%
B. 15.23%
C. 25.25%
D. 30.40%
E. 28.45%

Sol. Surface area of square is 4ᴨr2
Radius increases by 5%, so by successive increase formula
surface area changes by 5+5+((5*5)/100) = 10.25%
Total cost of painting = Surface area of sphere * Cost
So again by successive increase formula
change in cost of painting will be = 10.25+10+((10.25*10)/100).

10. 2 pipes A and B are such that A fills the bucket in 8 hrs and B empties the bucket in 4 hrs. If pipe A is opened first and after 2 hours B is also opened, in how much total time (in hours) will the bucket get empty ?

A. 5
B. 4
C. 3
D. 2
E. None of these

Sol. In 2 hrs, bucket filled by A is 2 * 1/8 = 1/4
After 2 hrs, bucket filled by A + B in 1 hr is 1/8 -1/4 = -1/8
This means after 2 hrs, 1/4th is full, now in next 1 hr 1/8 out of 1/4th already filled gets empty, so now 1/4 – 1/8 = 1/8 bucket is filled, in next 1 hr (A+B) empties this 1/8 too
So total 2 + 1 + 1 = 4.
For more Updates & Notifications for bank exams like SBI PO, SBI CLERK, IBPS PO, IBPS RRB, IBPS CLERK, IBPS SO, NIACL, SSC CHSL, SSC CGL etc. like our facebook page at :

0 comments:

Post a Comment

Popular Posts