IBPS RRB Mains 2016-17 (Important Questions-2)

IBPS RRB Mains 2016-17 (Important Questions-2)

1. The area of a rectangle is 336 square centimeter. Breadth of the rectangle is four times the radius of a circle. If the perimeter of the rectangle is 80 cm, then the radius of the circle is ?

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

Sol. Lets assume the length to be 'l'
And breadth to be 'b'.

According to the question,

l*b = 336

and

2(l+b) = 80

Solving , b = 12 cm

Thus, radius of the circle = b/4 = 12/4 = 3cm.

2. Three dices are rolled simultaneously then find the probability that none of them have all same numbers ?

A. 35/36
B. 1/36
C. 5/36
D. 1/216
E. None of these

Sol. Total number of cases = 6*6*6 = 216
No. of cases where all have same numbers = 6
Number of favorable cases = (216 - 6) = 210

Thus, required probability = 210/216 = 35/36.

3. A can do a piece of work in 10 days and B can do the same work in 15 days. They work together for 3 days then B left. Find in how many days A finish the remaining work ?

A. 4 
B. 3
C. 5
D. 6
E. None of these

Sol. In 1 day working together they finish, 1/10 + 1/15 = 1/6th work.
In 3 days they will finish, = 1*3/6 = 1/2th work
Work remaining, to be done by A alone is  = 1/2th work

A finihes the whole work in = 10 days
Thus, A will finish 1/2th work in = 10/2 = 5 days.

4. An item was sold for Rss 2750. But if it was sold for 10% less, it would still made 10% profit. What is the cost price of the item ?

A. 2250
B. 2350
C. 2450
D. 2150
E. None of these

Sol. 10% 0f 2750 = Rs 275
Selling Price = 2750 - 275 = Rs 2475

Cost price = (2475*100)/(100+10) = Rs 2250.

5. In an exam, 10% boys and 10% girls passed. Number of boys is 100 and number of girls is 90. What is the percentage of candidates passed ?

A. 90
B. 80
C. 10
D. 20
E. None of these

Sol. No. of boys = 10% of 100 = 10
No. of girls = 10% of 90 = 9

Total candidates passed = 10 + 9 = 19

Percentage Failed = (1 - 19/190)*100 = 90%.

Directions (6-10): What will come in place of (_) in the following number series ?

6. 11, 13, 17, 19, 23, 29, _

A. 31
B. 37
C. 33
D. 34
E. None of these

Sol. All are prime numbers. Next prime is 31.

7. 3, 7, 15, 31, 63, _

A. 125
B. 123
C. 127
D. 129
E. None of these

Sol. 3*2 + 1 = 7
     7*2 + 1 = 15
    15*2 + 1 = 31
    31*2 + 1 = 63
    63*2 + 1 = 127.

8. 12, 36, 18, 54, 27, _

A. 81
B. 80
C. 87
D. 82
E. None of these

Sol. 12*3 = 36
     36/2 = 18
     18*3 = 54
     54/2 = 27
     27*3 = 81.

9. 9, 28, 65, 126, 217, _

A. 344
B. 354
C. 364
D. 374
E. None of these

Sol. 2^3 + 1 = 9
     3^3 + 1 = 28
     4^3 + 1 = 65
     5^3 + 1 = 126
     6^3 + 1 = 217
     7^3 + 1 = 344

10. 6, 18, 72, 360, 2160, _

A. 15120
B. 14250
C. 16540
D. 18750
E. None of these

Sol. 3*2 = 6
     6*3 = 18
    18*4 = 72
    72*5 = 360
   360*6 = 2160
  2160*7 = 15120. 

Read More

IBPS RRB Mains 2016-17 (Important Questions)

IBPS RRB Mains 2016-17 (Important Questions)

                                                                                                                                                                                                                                                                                                                                                                                                                                                                              

Directions (1-5): What will come in place of (_) in the following number series :

1. 7, 20, 46, 85, _, 202

A. 136

B. 135

C. 137

D. 138

E. 141

Sol. 20 - 7 = 13

       46 - 20 = 26

       85 - 46 = 39

     

   Thus, 85 + 52 = 137.

2. 4, 13, 40, 121, _, 1093

A. 365

B. 364

C. 366

D. 368

E. None of these

Sol. 13 - 4 = 9

     40 - 13= 27

    121 - 40= 81

  

Thus, 121 + 243 = 364.

3. 17, 18, 14, 23, 7, _

A. 30

B. 32

C. 33

D. 36

E. 39

Sol. 18 - 17 = 1

     14 - 18 =-4

     23 - 14 = 9

      7 - 23 =-16

Thus, 7 + 25 = 32.

4. 2180, 2179, 2152, 2027, _

A. 1780

B. 1684

C. 1500

D. 1980

E. None of these

Sol. 2179 - 2180 =-1

     2152 - 2179 =-27

     2027 - 2152 =-125

Thus, 2027 - 343 = 1684.

5. 6, 10, 19, 32, 48, _

A. 64

B. 65

C. 66

D. 67

E. 68

Sol. 10 - 6 = 4

     19 - 10= 9

     32 - 19= 13

     48 - 32= 16

     9 - 4 = 5

    13 - 9 = 4

    16 - 13= 3

Thus, 16 + 2 = 18

Therefore, 48 + 18 = 66.

6. 5 years ago the age of A was one-third the age of B, and sum of their present ages is 70. Then find the age of A, 5 years after ?

A. 15

B. 20

C. 25

D. 30

E. None of these

Sol. Lets assume B's age 5 years ago be 'x'

Then A's age 5 years ago was x/3

According to the question, 

x + 5 + x/3 + 5 = 70

Solving, x = 45

Thus, A's present age = 45/3 + 5 = 20

A's age 5 years after = 20 + 5 = 25 years. 

7. If 40% of X is added to Y, the resultant is one-fifth more than Y. Calculate ratio X:Y ?

A. 2:3

B. 1:2

C. 3:4

D. 1:4

E. 1:5

Sol. According to the question, 

0.4X + Y = Y + Y/5

Solving, X/Y = 1/2

Thus, X:Y = 1:2.

8. Shiya's income is 25% more than Aman, and if each save 1/3 of their incomes, and total of their savings is Rs 11000. What is the difference in their incomes ?

A. 3500

B. 3666.67

C. 4666.67

D. 4500

E. None of these

Sol. Lets assume Aman's income be 'x'

Then, Shiya's income is 1.25x

According to the question,

1.25x/3 + x/3 = 11000

Solving, x = 14666.67

Difference in incomes = 0.25*14666.67 = Rs 3666.67 

9. In a lab, two chemical solutions Sulphuric Acid with 94% purity and Hydrochloric Acid with 96% purity are mixed resulting in 24 litres of mixture of 95% purity. How much is the quantity of the Sulphuric Acid in the resulting mixture ?

A. 12

B. 21

C. 20

D. 16

E. 18

Sol. 94x + (24 – x)* 96 = 24 * 95

2x = 24

x = 12.

10. A basket contains 5 red 4 blue 3 green balls. If three balls picked up random, What is the probability that at least one ball is blue ?

A. 41/55

B. 53/55

C. 47/55

D. 49/55

E. 31/55

Sol. Total Balls = 12

Other than blue = 8c3/12c3 = 14/55

Probability = 1-14/55 = 41/55.

Read More

IBPS RRB 2016-17 - Quant Quiz-21

1. In a School of X Students ,when Rahul was absent , the number of students could be divided into groups of 6 each If both Ramesh and Vikas were absent ,the remaining could be divided into groups of 6 each. What is the minimum Possible value of X ?

A. 63
B. 37
C. 69
D. 84
E. None of these

Sol. If 1 Student is absent the no. of students should be in the form of 6a + 1
7, 13, 19, 25, 31, 37, 43 …
If 2 Students are absent the no. of students should be in the form of 7b + 2
9, 16, 23, 30, 37, 44 …
So take the common term as 37.

2. Time taken by A and B together to complete a work is equal to the time taken by C and D together to complete that work .Also, time taken by B and C together to complete that work is two times the time taken by A and D together. If B alone can complete the work in 10 days and D alone can complete the work in 5 days, What is the ratio of efficiencies of A , B , C and D ?

A. 1:2:2:1
B. 2:1:1:2
C. 1:1:2:2
D. 2:1:1:1
E. None of these

Sol. P + Q = R + S
Also, Q + R = (P + S)/2
Put P and S value in above two equations
We have P – R = 1/10 & P = 2R
R = 1/10
P = 1/5
Ratio of Efficiencies = 1/5 : 1/10 : 1/10 : 1/5
2 : 1 : 1 : 2

3. In a race of 100 meter , R beats A by 40 meters , and A beats M by 40 meters .If R completes the race in 9 seconds after how much time does M complete the race ?

A. 14 sec
B. 15 sec
C. 25 sec
D. 5  sec
E. 3  sec

Sol. When Ram runs 100m Ananth runs 60 m
Ratio of speeds = 5 : 3
Ratio of speeds of Ananth and Madhavan = 5 : 3
Ratio of their three Speeds = 25 : 15 : 9
Time is inverse to speed , So
1/25 : 1/15 : 1/9
9 : 15 : 25
Madhavan finishes in 25 seconds

4. A car covers a certain distance in 12 hours at a uniform speed .If the speed is increased by 6 km/hr then the same distance would get covered in 7.5 hours. What is the distance covered ?

A. 140 km
B. 150 km
C. 120 km
D. 160 km
E. None of these

Sol. D/7.5 – D/12 = 6
Solve D = 120.

5. Ravi and Arun are partners who share profit and loss in the ratio of 3:1.They agree to take Manish into Partnership on 1/5 th share of profit. The new sharing will be :

A. 4:5:6
B. 3:1:2
C. 5:6:1
D. 3:1:1
E. None of these

Sol. Total profit 100%
Manish = 1/5 = 100/5 = 20%
Ravi and Arun 3/4 : 1/4 remaining 80%
3 * 20 : 1 * 20 : 20
3 : 1 :1

6. 124, 228, 436, _, 1684, 3348

A. 944
B. 852
C. 872
D. 444
E. None of these

Sol. The series is
× 2 - 20, ×2 - 20, × 2 - 20, ….

7. 53, 75, 107, 149, ?, 263

A. 199
B. 200
C. 201
D. 195
E. None of these

Sol. The series is
+ 22, +32 , +42, +52, +62 ….

8. 10, 17, 48, 165, _, 3475, 20892

A. 688
B. 712
C. 848
D. 918
E. None of these

Sol. The series is
×1 + 7, ×2 + 14, × 3 + 21, ×4 + 28 ….

9. 15, 17, 38, 120, 488, _, 14712

A. 2450
B. 2650
C. 2850
D. 2950
E. None of these

Sol. The series is

× 1 + 2, ×2+ 4 , ×3 + 6, × 4 + 8, × 5 + 10

10. 1, 6, 19, _, 85, 146, 231

A. 46
B. 44
C. 48
D. 65
E. None of these

Sol.  6 - 1 = 5
     19 - 6 = 13
     44 - 19= 25
     85 - 44= 41
    146 - 85= 61
    231 - 146=85

     13 - 5 = 8
     25 - 13= 12
     41 - 25= 16
     61 - 41= 20
     85 - 61= 24. 

Read More

IBPS RRB 2016-17 - Quant Quiz-20

1. Three friends X, Y, Z invested in a business in the ratio of 4:5:6. After 6 months Z withdraw half of his capital. If the sum invested by X is 36000, then the profit earned by Z out of the total profit of 60000 ?

A.20000
B.30000
C.40000
D.50000
E.None of these

Sol. Sum invested by X = 4x = 36000.
X = 9000
Investment made by X, Y , Z – 36000, 45000, 54000
Ratio in which the profit will divide- 36000*12 : 45000*12 :(54000*6 + 27000*6)
i.e 4:5:9. So Z share = (9/18)*60000 = 30000

2. Twice the speed downstream is equal to the thrice the speed upstream, the ratio of speed in still water to the speed of the current is: 

A. 1 : 5 
B. 5 : 1
C. 1 : 3 
D. 2 : 3
E. None of these

Sol. Let, speed in still water = x Km/h.
Speed of current = y Km/h.
Speed downstream = (x + y) Km/h.
Speed upstream = (x – y) Km/h.
 2(x + y) = 3(x – y)
 x = 5 y
or, x/y=5/1 or 5 : 1.

3. Karan decided to donate 5% of his salary. On the day of donation, he changed his mind and donated Rs. 1687.50 which was 75% of what he had decided earlier. How much is Karan's salary? 

A. Rs. 37.500   
B. Rs. 45,000 
C. Rs. 33,750   
D. Cannot be determined  
E. None of these  

Sol. Karan donated = (5*75)/100=15/4%
Then, 15/4 % = 1687.50
Karan's salary = [(1687.50*4)/15]*100=45000

4. The present ages of three persons are in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).

A. 8, 20, 28 
B. 16, 28, 36 
C. 20, 35, 45
D. 25, 30, 40 
E. None of these

Sol. Let their present ages be 4x, 7x and 9x years respectively.
Then, (4x – 8) + (7x – 8) + (9x – 8) = 56 => 20x = 80 => x = 4.
Their present ages are 16 yrs, 28 yrs. and 36 yrs. respectively.

5. The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages is 34. Find the present age of father.

A. 32 years 
B. 36 years 
C. 38 years
D. 40 years 
E. 39 years

Sol. Let the ages of father and son be x and (45 – x) years respectively.
Then, (x – 5) (45 – x – 5) = 34
(x – 5) (40 – x) = 34 => x2 – 45x + 234 = 0
(x – 39) (x – 6) = 0 => x = 39 or x = 6.
Father’s age = 39 years and son’s age = 6 years

6. A box contains 3 red, 6 blue,2 green and 4 yellow balls. If two marbles are picked randomly then the probability that either both are red or both are green is ?

A. 3/5   
B. 4/105   
C. 2/7
D. 5/91   
E. None of these

Sol. P(E) = (3C2 + 2C2)/15C2 = 4/105.

7. The simple interest on a sum of money is 4/9 of the principal and the number of years is equal to the rate per cent per annum. The rate per annum is: 

A. 5%
B. 6(2/3)%
C. 6%
D. 7(1/5)%
E. None of these

Sol.Let the sum of money be P
I = (P×R×T)/100
(4/9)P = (P×R×T)/100
R = √(400/9) = 20/3 = 6(2/3)%.

8. What would be the compound interest accrued on an amount of 8000 at the rate of 15% per annum in three years ?

A. 4283 
B. 4051 
C. 4167
D. 4325 
E. None of these

Sol. CI =8000[(1+15/100)^3-1]  
= 8000[(115/100)^3-1] = 4167.

9. The circumference of base of a circular cylinder is 6π cm. The height of the cylinder is equal to diameter of the base. Maximum amount of water that can be poured in the cylinder is

A. 54π cm3    
B. 36π cm3
C. 0.054π cm3    
D. 0.54π cm3
E. None of these 

Sol. Given 2πr = 6π
r = 3
h = 2*3 = 6
V = πr2h = π×3×3×6 = 54 π.

10. Two vessels contain milk and water in the ratio 3 : 2 and 7 : 3. Find the ratio in which the contents of the two vessels have to be mixed to get a new mixture in which the ratio of milk and water is 2 : 1.   

A. 2 : 1
B. 1 : 2 
C. 4 : 1 
D. 1 : 4
E. None of the above

Sol. = (21 - 20)/30
= (10 - 9)/15
= 1/30 = 1/15
Required ratio = 1/30 : 1/15 = 1:2.

Read More

IBPS RRB 2016-17 - Quant Quiz-19

1. 3, 8, 15, 24, 35, _

A. 47
B. 48
C. 49
D. 50
E. None of these

Sol. 1*2 + 1 = 3
     2*3 + 2 = 8
     3*4 + 3 = 15
     4*5 + 4 = 24
     5*6 + 5 = 35
     6*7 + 6 = 48

2. 2, 3, 10, 15, 26, _

A. 36
B. 35
C. 34
D. 32
E. None of these

Sol. 1^2 + 1 = 2
     2^2 - 1 = 3
     3^2 + 1 = 10
     4^2 - 1 = 15
     5^2 + 1 = 26
     6^2 - 1 = 35

3. 5, 11, 19, 29, 41, _

A. 46
B. 48
C. 55
D. 57
E. None of these

Sol. 1 + 2^2 = 5
     2 + 3^2 = 11
     3 + 4^2 = 19
     4 + 5^2 = 29
     5 + 6^2 = 41
     6 + 7^2 = 55

4. 6, 12, 18, 24, 30, _

A. 34
B. 35
C. 36
D. 38
E. None of these

Sol. 5*1 + 1 = 6
     5*2 + 2 = 12
     5*3 + 3 = 18
     5*4 + 4 = 24
     5*5 + 5 = 30
     5*6 + 6 = 36

5. 6, 10, 4, 12, 2, _

A. 16
B. 15
C. 14
D. 18
E. None of these

Sol. 8 - 2*1 = 6
     6 + 2*2 = 10
    10 - 2*3 = 4
     4 + 2*4 = 12
    12 - 2*5 = 2
     2 + 2*6 = 14

6. There are 2 schemes A and B, both have 12% interest rate. A person invests in them in the ratio of 3:4. After 3 years, the difference between interest from A and interest from B is Rs 360, calculate the amount invested in scheme A ?

A. 4000
B. 3000
C. 2000
D. 1000
E. None of these

Sol. Lets assume investment by A to be '3x'.
Assume invested by B to be '4x'.

S.I. from A = (3x*3*12)/100 = 1.08x
S.I. from B = (4x*3*12)/100 = 1.44x

Difference = 1.44x - 1.08x = 0.36x
Thus, 0.36x = 360
x = Rs 1000

Investment by A = 3*1000 = Rs 3000.

7. The height of 5 boys is recorded as, 146 cm, 154 cm, 164 cm, 148 cm and 158 cm. What is the average height of all these boys ?

A. 158
B. 156
C. 154
D. 152
E. None of these

Sol. Average height = (146=154+164+148+158)/5 = 154.

8. Pravin purchased 25 kg of rice at the rate of Rs 45 per kg and 12 kg of pulses at the rate of Rs 28 per kg. What is the total amount that he paid to the shopkeeper ?

A. 1461
B. 1416
C. 1561
D. 1516
E. None of these

Sol. Total amount paid = 25*45 + 12*28 = Rs 1461.

9. The perimeter of a rectangle is 60 cm and its breadth is 12 cm. What is the area of the rectangle ?

A. 220
B. 240
C. 180
D. 216
E. None of these

Sol. Perimeter = 2(l+b)
As given, 2(l+b) = 60
Also given, b = 12
Thus, 2(l+12) = 60
l = 18
Area = 18*12 = 216.

10. If the three fourth of a number is subtracted from the number; the value so obtained is 163. What is that number ?

A. 652
B. 556
C. 672
D. 572
E. None of these

Sol. x - 3x/4 = 163
x/4 = 163
x = 652.

Read More

IBPS RRB 2016-17 - Quant Quiz-18

1. Rakesh invested Rs 20,000 in a scheme exactly 4 years ago. The value of investment increased by 15% during the 1st year, increased by 10% during the 2nd year, decreased by 5% during the 3rd year and decreased by 10% during the 4th year. What is the value of investment today ?

A. 21,600
B. 21,631.5
C. 22,000
D. 22,600
E. None of these

Sol. 20000*1.15*1.1*0.95*0.9 = Rs 21631.5

2. Tina bought 25 tabs and phones for Rs 205000. She sold 80% of tabs and 12 phones for a profit of Rs 40,000. Each tab was marked up by 20% over cost and each phone was sold at a profit of Rs 2000. The remaining tabs and 3 phones could not be sold. What is tina's overall profit/loss ?

A. 2000
B. 1000
C. 1500
D. 2500
E. None of these

Sol. Total no. of phones = 12 + 3 = 15
     Total no. of tabs = 25 - 15 = 10

Total CP = Rs 205000
Now, cost of 80% of the goods,
= 0.8*205000
= Rs 164000

Total amount recovered (or SP)
= 164000 + 40000
= Rs 204000

Loss = 205000 - 204000 = Rs 1000

3. Sum of CP's of 2 cows is Rs 13,000. Both the cows are sold at a profit of 20% and 40% respectively with their SP's being the same. What is the difference of CP's of both the cows ?

A. 1500
B. 1200
C. 1000
D. 2000
E. None of these

Sol. Lets assume the CP's of two cows will be x and y.
x + y = 13000
SP of 1st cow = 1.2x  ( as 20% profit)
SP of 2nd cow = 1.4y  ( as 40% profit)
As both are same,
1.2x = 1.4y
x = 7y/6
Thus,
7y/6 + y = 13000
13y/6 = 13000
y = 6000
x = 7000
Thus, difference = 7000 - 6000 = Rs 1000

4. A tank has 2 pipes. 1st pipe can fill in 45 min. and 2nd pipe can empty in 1hr. In what time will the empty tank be filled if the pipes be opened one at a time in alternate minutes ?

A. 5hr 53min
B. 4hr 42min
C. 3hr 13min
D. 4hr 43min
E. None of these

Sol. Part of tank that will empty in 2min = 1/45 - 1/60 = 1/180
Part of the tank (1 - 1/45) = 44/45 will be filled in = 44*180*2/45 = 352min.
1/45 part of the tank will be filled in 1 min.
Total time = (352+1) = 353 min = 5hr 53min

5. A person used to drawn out 20% of the honey from the full jar and replaced it with sugar solution. He has repeated the same process 4 times and thus there was only 512 gm of honey left in the jar, the rest part of the jar was fiilled with sugar solution. The initial amount of honey in the jar was :

A. 1 kg
B. 1.5 kg
C. 2 kg
D. 2.5 kg
E. None of these

Sol. Let initial amount of honey be 'k' gm.
Thus, 512 = k(1-1/5)^4
512 = k(4/5)^4
k = 1250 gm
Therefore, initial amount of hney was = 1250/1000 = 1.25 kg.

6. A river is flowing at a speed of 10 km/h in a particular direction. Mr.Suresh, who can swim at a speed of 15 km/h in still water, starts swimming along the direction of flow of the river from points A and reaches another point B which is at a distance of 50 km from the starting point A. On reaching point B, he turns back and from starts swimming against the direction of flow of the river and stops after reaching point A. The total time taken by Suresh to complete his journey is ?

A. 11
B. 12
C. 9
D. 8
E. None of these

Sol. Speed of the stream = 10kmph
Downstream speed = 15 + 10 = 25kmph
Upstream speed = 15 – 10 = 5kmph
Total time taken by Suresh = (50/25 + 50/5) = 12h

7. Three persons A, B and C started a business together. They invest Rs.40000, Rs.24000 and Rs.30000 respectively in the beginning. After 4 months. B took out Rs. 4000 and C took out Rs. 10000. They get a profit of Rs. 25400 at the end of the year. B’s share in the profit is ?

A. 6500
B. 6400
C. 5600
D. 5200
E. None of these

Sol. 40000 * 12 : 24000 * 4 + 20000 * 8 : 30,000 * 4 + 20,000 * 8
60 : 32 : 35
32/127 * 25400 = 6400

8. Two cars X and Y starts at the same time from D and A respectively which is 200 km apart, If the two cars travel in opposite directions they meet after 1 hour and if they travel in same direction from D and A then the car which starts at D meets the another car after 5 hours, What is the speed of car starting from D ?

A. 120
B. 100
C. 180
D. 125
E. None of these

Sol. Relative speed = x+y
total distance travelled = 200 km
time taken = 1 hour
x + y = 200 —(1)
Relative speed of the faster car = x – y
total distance travelled = 200 km
time taken = 5 hour
200 = 5(x-y)
40 = x – y —(2)
From (1) and (2) 2x = 240
x = 120
Speed of car starting from D = 120km/hr.

9. Ramya’s present age is four times her daughter, Veena’s present age and two-thirds of her mother, Rani’s present age. The total of the present ages of all of them is 132 years. What is the difference between Ramya’s and her mother Rani’s present age ?

A. 28
B. 24
C. 20
D. 26
E. None of these

Sol. Ramya’s present age – x
x + (x / 4) + (3 / 2) x = 132
11x = 132 * 4
x = 48
Ramya’s mother age = 3/2 * 48 = 72
Difference = 72 – 48 = 24 years.

10. The circumference of a semicircle of area 1925 sq. cm is equal to the breadth of a rectangle. If the length of the rectangle is equal to the perimeter of a square of side 48 cm. What is the perimeter of the rectangle in cm ?

A. 745
B. 744
C. 732
D. 736
E. None of these

Sol. (22/7 *r *r) /2 = 1925
r* r = 1225
r = 35
Circumference of semicircle = 22/7 r + 2 r = 22/7 * 35 + 2 * 35 =180
Breadth of rectangle = 180 cm
Length of rectangle = perimeter of square = 4 * 48 = 192 cm
Perimeter of rectangle = 2 (180 + 192) = 744 cm

Read More

IBPS RRB 2016-17 - Quant Quiz-17

1. 8, 16, 30, 52, 84, _

A. 120
B. 124
C. 128
D. 132
E. None of these

Sol. 8 + 3^2 - 1 = 16
    16 + 4^2 - 2 = 30
    30 + 5^2 - 3 = 52
    52 + 6^2 - 4 = 84
    84 + 7^2 - 5 = 128

2. 11, 13, 29, 91, 369, _

A. 1851
B. 1861
C. 1841
D. 1871
E. 1881

Sol. 11*1 + 2 = 13
     13*2 + 3 = 29
     29*3 + 4 = 91
     91*4 + 5 = 369
    369*5 + 6 = 1851

3. 4, 6, 12, 24, 44, _

A. 64
B. 74
C. 84
D. 94
E. None of these

Sol. 4 + 1*2 = 6
     6 + 2*3 = 12
    12 + 3*4 = 24
    24 + 4*5 = 44
    44 + 5*6 = 74

4. 8, 9, 17, 44, 108, _

A. 233
B. 234
C. 235
D. 236
E. 237

Sol. 8 + 1^3 = 9
     9 + 2^3 = 17
    17 + 3^3 = 44
    44 + 4^3 = 108
   108 + 5^3 = 233

5. 25, 313, 457, 529, 565, _

A. 577
B. 583
C. 593
D. 589
E. None of these

Sol. 25 + 9 * 32 = 313
    313 + 9 * 16 = 457
    457 + 9 * 8 = 529

Directions (6-10): Read the information given and answer the following questions accordingly.

Not surprisingly the growth of the hotel industry is driven by the increase in the number of people using hotels and the increase in per person use of the hotel. In 2004, it is expected that there will be 200 million hotel users in India or about 20 per cent of the population will generate Rs. 50 billion in hotel revenues. Industry revenues should expand from Rs. 50 billion to Rs. 150 billion by 2008, while the number of users should grow to over 560 million or to about half the population of India in the same period. 

6. What is the estimated population of India in 2004? 

A. 98 crore 
B. 100 crore 
C. 110 crore 
D. 115 crore 
E. None of these

Sol. 200 million = 20% of population
⇒ Population = 200 × 5 = 1000 million = 100 crore

7. What will be the simple average growth rate of population of India in the given period 2004-2008?  
A. 2%
B. 3%
C. 4%
D. 4.5%
E. None of these

Sol. 2004 population = 1000 million
Population in 2008 or after 4 years = 560 × 2 = 1120 million
∴ Growth rate = (120 × 100)/(1000 × 4) = (12/4)% = 3% per annum simple growth rate.

8. What will be the growth in percentage of users in India by 2008?  

A. 100%
B. 150%
C. 180%
D. 200%
E. None of these

Sol. Hotel users in 2004 = 200 million
Hotel users in 2008 = 560 million
∴ Growth in percentage =(560 - 200)/200=360/200=180%

9. What will be the percentage growth of the revenues of the hotel industry in the given period?  

A. 200%
B. 230%
C. 260%
D. 300%
E. None of these

Sol. Total revenue in 2004 = 50 billion
Total revenue in 2008 = 150 billion
∴ Growth in percentage =(150 - 50)/50×100=100/50=200%  

10. It is believed that if 50% of the population of any country can afford hotel-use, it is economically developed. Can we say that India will be a developed country by 2007?  

A. Yes 
B. No 
C. Cannot say 
D. Data inadequate 
E. None of these

Sol. By 2008 half or 50% of the India population will be using hotels.
We do not have any information about 2007. Hence the data is inadequate. 

Read More

IBPS RRB 2016-17 - Quant Quiz-16

1. A shopkeeper bought 25 kg of Rice at rate of Rs.40 per kg. He sold 60% of the total quantity at the rate of Rs.50 per kg. Exactly, at what price per kg should he sell the remaining quantity to make 50% overall profit ?

A. 85
B. 80
C. 75
D. 70
E. None of these

Sol. CP of Rice = 25 * 40 = 1000
60% of 25 kg = 15 kg
SP of 15kg = 15 * 50 = 750
150/100 * 1000 = 1500
Difference = 1500 – 750 = 750
SP of 10kg = 750
SP of 1kg = 75

2. The smallest side of a right-angled triangle is 8 cm less than the side of a square of perimeter 56 cm, The second largest side of the right-angled triangle is 4 cm less than the length of the rectangle of area 96 sq cm and breadth 8 cm. What is the largest side of the right-angled triangle ?

A. 20 cm
B. 15 cm
C. 10 cm
D. 25 cm
E. None of these

Sol. Smallest side of a right-angled triangle = 1/4 (Perimeter of Square) – 8
= 1/4 (56) – 8 = 14 – 8 = 6
second largest side of the right-angled triangle = 96/8 – 4 = 8
largest side of the right-angled triangle = sq.root of (6 * 6 + 8* 8) = 10 cm

3. X, Y and Z started a business with their investments in the ratio 1:3:8. After 6 months, X invested the half amount more as before and Y invested twice the more as before, while Z withdraws 1/4 th of their investments. Find the ratio of their profits at the end of the year is ?

A. 5:24:28
B. 3:5:10
C. 1:2:3
D. 3:7:12
E. None of these

Sol. Investment of X for 1 year = x*6 + (x + x/2)6 = 15x
Investment of Y for 1 year = 3x*6 + (3x + 6x)6 = 72x
Investment of Z for 1 year = 8x*6 + (8x – 8x/4)6 = 84x
Ratio = 15x : 72x : 84x = 5:24:28

4. A butler stole wine from a butt of sherry which contained 80% of spirit and he replaced it by wine containing only 32% of spirit. The butt was now only 48% strong. How much of the butt did he steal ?

A. 1/3
B. 2/3
C. 1/5
D. 2/5
E. None of these

Sol. 32 ———- 80
48
32 ———- 16
Ratio = 2:1
butler stole 2/3 of butt.

5. The ratio of the adjacent angles of a parallelogram is 7 : 8. Also, the ratio of the angles of quadrilateral is 5 : 6 : 7 : 12. What is the sum of the smaller angle of the parallelogram and the second largest angle of the quadrilateral ?

A. 168
B. 188
C. 224
D. 216
E. None of these

Sol. Sum of the adjacent angles of Parallelogram = 180°
Smaller angle of Parallelogram = 7/15 * 180 = 84°
Second largest angle of the quadrilateral = 7/30 * 360 = 84°
Sum = 84° + 84° = 168°

6. Shopkeeper purchased some goods for Rs.900 and sold one-third of the goods at a loss of  what 12%, then at gain % should the remainder goods he sold to gain 18% profit on the whole transaction ?

A.31%
B.26%
C.33%
D.18%
E.None of these

Sol. 1/3rd at 12% loss = 900/3 = (300*88)/100 = 264
(900*18)/100 = 108
600 + 162 + 36 = 798
(198/600)*100 = 33%

7. A dealer marked the price of an item 20% above cost price. He allowed two successive discounts of 10% and 25% to a customer. A a result he incurred a loss of Rs.684. At what price did he sell the item to the customer ?

A.1670
B.2768
C.1817
D.2916
E.None of these

Sol. CP = 100
MP = 120
120*90/100 = 108; 108*75/100 = 81
Loss = 100 – 81 = 19%
CP = 100/19*684 = 3600
SP = 3600*81/100 = 2916

8. A man went to shop to buy mobile for Rs.3500, the rate of sales tax being 8%. He tell the shopkeeper to reduce the price of mobile to such an extent that has to pay Rs.3500, inclusive of all taxes. Find the reduction price.

A.Rs.210
B.Rs.260
C.Rs.320
D.Rs.140
E.None of these

Sol. (3500*100)/108 = 3240
3500-3240 = 260

9. An article passing threw two hands, is sold at profit of 40 % at the original price , if the first dealer makes a profit of 15 % , then the profit made by second dealer.

A.25%
B.26%
C.18%
D.22%
E.None of these

Sol. Total Profit = [P1 + P2 +(P1*P2/100)] 40 = 15 + P2 + (15*P2)
25-p2 = 15P2/100
2500 = 15p2+100p2
P2 = 2500/115 = 21.7% = 22%

10. What is the maximum % discount that a man can offer on her MP so that he ends up selling at no profit or loss, if he had initially marked his goods up by 35% (approximately)?

A.20%
B.32%
C.27%
D.30%
E.None of these

Sol. 35-x-(35*x/100) = 0
3500/100 –(100x+35x/100) = 0
35 = 135/100
X = 35*100/135 = 25.9 = 26%

Read More

IBPS RRB 2016-17 (Questions asked in 4th Nov. shift)

Directions(1-5): In each question two equations are given, find x and y and give the answer:

A. x > y
B. x < y
C. x >= y
D. x =< y
E. x = y or relation can not be established.

1. (1) 8x^2 - 106x + 323 = 0
   (2) 12y^2 - 109y + 247 = 0

Answer. x = 17/2 , 19/4
        y = 13/3 , 19/4
Thus, x >= y.

2. (1) 4x + 7y = 42
   (2) 3x - 11y = -1

Answer. x = 7
        y = 2
Thus, x > y.

3. (1) 9x^2 - 29x + 22 = 0
   (2) y^2 - 7y + 12 = 0 

Answer. x = 2 , 11/9
        y = 3 , 4
Thus, x < y.

4. (1) 3x^2 - 4x - 32 = 0
   (2) 2y^2 - 17y + 36 = 0

Answer. x = 4 , -8/3
        y = 4 , 9/2
Thus, x =< y

5. (1) 3x^2 - 19x - 14 = 0
   (2) 2y^2 + 5y + 13 = 0

Answer. x = 7 , -2/3
        y = -3/2 , -1
Thus, x > y.

6. A merchant purchases a wrist watch for Rs 450 and fixes its list price in such a way that after allowing a discount of 10%, he earns a profit of 20%. Find the list price ?

A. 550
B. 600
C. 500
D. 650
E. None of these

Sol. C.P. of wrist watch = Rs 450
     Discount = 10%
     Profit earned = 20%
Lets asssume list price to be = 'x'
Then,
450*(120/100) = x*(90/100)
Solving, x = Rs 600.

7. An article was sold at profit of 12%. If the C.P. would be 10% less and selling price is Rs 5.75 more then there would be 30% profit.Then at what price it should be sold to make a profit of 20% ?

A. Rs 136         
B. Rs 138
C. Rs 140
D. Rs 135
E. None of these

Sol. Lets assume C.P. be 'x'
Then, S.P. will be = 1.12x ( as 12% profit)

Now, C.P. = 0.9X (as C.P. is reduuced by 10%)
     S.P. = 1.12x + 5.75
Now as profit is 30%,
(130/100)*0.9x = 1.12x + 5.75
Solving, x = Rs 115

Now at 20% profit S.P. will be = (120/100)*115 = Rs 138.

8. Th simple interest on a sum of money is 8/25 of sum. If the number of years numerically is half the rate of interest then what is the rate of interest ?

A. 5
B. 4
C. 6
D. 8
E. None of these

Sol. Given, S.I. = (8/25)P
            Time = t = R/2

Now, S.I.= PRT/100
     (8/25)P = [P*R*(R/2)]/100
Solving, R = 6%.

9. Three glasses of equal volume contain acid mixed with water. The ratio of acid and water are 2:3, 3:4 and 4:5 respectively. Contents of these glasses are poured in a large vessel. The ratio of acid and water in the large vessel is :

A. 417:564
B. 401:544
C. 407:560
D. 411:540
E. None of these

Sol. Glass 1, acid = 2x ; water = 3x
     Glass 2, acid = 3y ; water = 4y
     Glass 3, acid = 4z ; water = 5z

New Glass composition, acid:water = (2x+3y+4z):(3x+4y+5z)
Also as the volumes of glasses are equal thus,
2x+3x = 3y+4y
5x = 7y
y = 5x/7
Also, 2x+3x = 4z+5z
      5x = 9z
      z = 5x/9

Putting these values above, 401:544.

10. The average weight of 36 students is 50 kg. If was found later that the weight of one of the student was misread as 73 kg, whereas his actual weight was 37 kg. Find the correct average weight(in kg).

A. 50
B. 49
C. 51
D. 47
E. 48

Sol. Sum of the weights/36 = 50
     Sum of the weights = 50*36 = 1800

     Correction in sum of the weights = 37 - 73 = -36
     Correct sum of the weights = 1800 - 36 = 1764

Thus, correct average is = 1764/36 = 49.

11. Train takes 4 seconds to pass a man. Another train travelling in opp. direction of same length takes 5 seconds to pass the man. Find the time taken by both trains to cross each other ?

A. 51/9 sec
B. 40/9 sec
C. 42/9 sec
D. 41/9 sec
E. None of these

Sol. Length of trains = x meter
     Speed of 1st train = x/4 m/sec
     Speed of 2nd train = x/5 m/sec

Relative speed = x/4 + x/5 = 9x/20 m/sec
Required Time = 2x/(9x/20) = 40/9 seconds.

12. One third of a certain journey is covered at the rate of 25 km/hr, one fourth at the rate of 30 km/hr and the rest at 50 km/hr. Find the average speed of the whole journey ?

A. 100/3 
B. 35
C. 30
D. 37
E. None of these

Sol. Lets assume the total distance to be 'd'.
The, average speed = d/[(d/3*25)+(d/4*30)+(5d/12*50)] = 100/3 km/hr.

13. 8 men and 6 women can complete a work in 6 days. 1 man works twice as much as 1 woman in a day. 8 men and 4 women started working, after 2 days 4 men left and 4 new women joined. In how many more days will the work be completed ?

A. 6
B. 7
C. 5
D. 4
E. None of these

Sol. (8M + 4W)6 = T
Also, M = 2W
Thus, (16W + 4W)*6 = T
      120W = T
and   60M = T
Now, (8M + 4W)*2 + (4M + 8W)*t = T
     (8M + 2M)*2 + (4M + 4M)*t = 60M
     20 + 8t = 60
     t = 5 days.

14. A motor boat travelling at the same speed can cover 25 km upstream and 39 km downstream in 8 hrs. At the same speed, it can cover 35 km upstream and 52 km downstream in 11 hrs. Find the speed of the stream ?

A. 5
B. 6
C. 4
D. 3
E. None of these

Sol. Lets assume upstream speed to be 'x' km/hr
     And downstream speed to be 'y' km/hr
Then, 25/x + 39/y = 8
And, 35/x + 52/y = 11
Solving these two,we get,
x = 5km/hr
y = 13 km/hr

Thus, speed of stream = (y - x)/2 = (13 - 5)/2 = 4 km/hr.

15. In a lab, 2 bottles contain mixture of acid and water in the ratio of 2:5  in 1st and 7:3 in 2nd. Find the ratio in which these two should be mixed so that the new ratio will become 2:3 ?

A. 21:8
B. 15:8
C. 9:8
D. 19:8
E. None of these

Sol. Using the allegation method, (3/10):(4/35) = 21:8.  

Read More

IBPS RRB 2016-17 (Questions asked in 4th Nov. shift) - Number Series

Directions(1-10): Find the missing numbers in the given series:

1. 2, 4, 11, 37, 153, _

A. 671
B. 771
C. 541
D. 877
E. None of these

Sol. 2*1 + 2 = 4
     4*2 + 3 = 11
    11*3 + 4 = 37
    37*4 + 5 = 153
   153*5 + 6 = 771  

2. 3, 10, 22, 39, 61, _

A. 86
B. 84
C. 83
D. 88
E. None of these

Sol. 3 + 7 = 10
    10 + 12= 22
    22 + 17= 39
    39 + 22= 61
    61 + 27= 88

3. 6, 6, 8, 14, 26, _

A. 42
B. 44
C. 46
D. 40
E. None of these

Sol. 6 + 1^2 - 1 = 6
     6 + 2^2 - 2 = 8
     8 + 3^2 - 3 = 14
    14 + 4^2 - 4 = 26
    26 + 5^2 - 5 = 46

4. 1, 3, 9, 31, 129, _

A. 560
B. 651
C. 543
D. 622
E. None of these

Sol. 1*1 + 2 = 3
     3*2 + 3 = 9
     9*3 + 4 = 31
    31*4 + 5 = 129
   129*5 + 6 = 651

5. 5, 11, 24, 44, 71, _

A. 102
B. 105
C. 112
D. 101
E. None of these

Sol. 5 + 6 = 11
    11 + 13= 24
    24 + 20= 71
    71 + 27= 105

    13 - 6 = 7
    20 - 13= 7
    27 - 20= 7

6. 3, 8, 15, 26, 39, _

A. 64
B. 56
C. 42
D. 48
E. None of these

Sol. 8 - 3 = 5
    15 - 8 = 7
    26 - 15= 11
    39 - 26= 13
    39 + (13+4) = 56

7. 9, 14, 21, 32, 45, _

A. 65
B. 60
C. 55
D. 70
E. None of these

Sol. 14 - 9 = 5
     21 - 14= 7
     32 - 21= 11
     45 - 32= 13
     45 + 15 = 60.

8. 10, 5, 5, 7.5, 15, _

A. 40
B. 37.5
C. 38
D. 45
E. None of these

Sol. 10*0.5 = 5
      5*1   = 5
      5*1.5 = 7.5
     7.5*2  = 15
     15*2.5 = 37.5

9. 6, 11, 18, 29, 46, _

A. 66
B. 71
C. 80
D. 95
E. None of these

10. 7, 15, 28, 59, 114, _

A. 225
B. 228
C. 233
D. 240
E. None of these

Sol. 7*2 + 1 = 15
    15*2 - 2 = 28
    28*2 + 3 = 59
    59*2 - 4 = 114
   114*2 + 5 = 233  

Read More

IBPS RRB 2016-17 - Quant Quiz-15

1. In a family, father’s age is twice that of son's age. Father is 10 years older than mother. Daughter is 20 years younger than his mother and 5 years younger than his brother. What is the age of the father?  

A. 52 years 
B. 50 years 
C. 58 years 
D. 55 years 
E. None of these

Sol. Le the age of father be 'x'
Then age of son will be = x/2
Age of mother will be = (x-10)
Age of daughter will be = (x-10-20) = (x-30)
Also given,
x/2 - 5 = x - 30
x = 50 yrs.

2. After replacing an old member by a new member, it was found that the average age of five members of a club is same as it was 3 years ago. The difference between the ages of the replaced and the new members is: 

A. 2 years 
B. 4 years 
C. 8 years 
D. 15 years
E. None of these

Sol. Increase in ages of five members in 3 years = 3*5 = 15 yrs
Since the avg. age remains same, therefore required difference = 15 yrs.

3. The ratio of the present age of X to that of Y is 3 : 11. Y is 12 years younger than Z. Z’s age after 7 years will be 85 years. What is the present age of X’s mother, who is 25 years older than X ?


A. 43 years
B. 67 years 
C. 45 years 
D. 69 years 
E. None of these 

Sol. Present age of Z = 85 - 7 = 78
     Present age of Y = 78 - 12 = 66
     Present age of X = (3/11)*66 = 18
     Present age of X's mother = 18 + 25 = 43 yrs.

4. At present, Tina is eight times her daughter’s age. 8 years from now, the ratio of the ages of Tina and her daughter will be 10 : 3. What is Tina’s present age?  

A. 32 years 
B. 40 years  
C. 36 years 
D. Cannot be determined 
E. None of these

Sol. Given T = 8D
where T is the age of tina
And D is the age of daughter
Also, (8D+8)/(D+8)=10/3
Solving, we get D=4
Tina's present age = 8D = 8*4 = 32 yrs.

5. The age of the father is 30 years more than the son’s age. Ten years hence, the father’s age will become three times the son’s age that time. What is the son’s present age in years?   

A. Eight  
B. Seven 
C. Five 
D. Cannot be determined 
E. None of these 

Sol. Let the son's present age be x yrs.
Then the father's present age is (x+30) yrs.
Father's age after 10 yrs = (x+40) yrs
Son's age after 10 yrs = (x+10) yrs
According to the question:
(x+40)=3(x+10)
x = 5 yrs

Directions(6-10): Find the missing number in the given series:

6. 4, 2, 8, -4, 16, _

A. -10
B. -12
C. -14
D. -8
E. None of these

Sol. 4 - (1*2) = 2
     2 + (2*3) = 8
     8 - (3*4) =-4
    -4 + (4*5) =16
    16 - (5*6) =-14

7. 1, 6, 9, 44, 85, _

A. 146
B. 136
C. 126
D. 116
E. 156

Sol. 1 + (1^2 + 2^2) = 6
     6 + (2^2 + 3^2) = 19
    19 + (3^2 + 4^2) = 44
    44 + (4^2 + 5^2) = 85
    85 + (5^2 + 6^2) = 146

8. 36, 202, 44, 290, _

A. 52
B. 42
C. 48
D. 50
E. 56

Sol. 10^2 - 8^2 = 36
     11^2 + 9^2 = 202
     12^2 - 10^2 = 44
     13^2 + 11^2 = 290
     14^2 - 12^2 = 52      

9. 9 16 44 107 ?

A. 362
B. 288
C. 282
D. 364
E. None of these

Sol. 9 + 2³ – 1 = 16
16 + 3³ + 1 = 44
44 + 4³ – 1 = 107; 107 + 5³ + 1 = 233

10. 13 17 33 97 ? 1377

A. 332
B. 353
C. 388
D. 396
E. 340
E. 233  

Sol. 13 + 2² = 17; 17 + 4² = 33; 33 + 8² = 97….

Read More

IBPS RRB 2016-17 - Quant Quiz-14

1. Find the number of ways of distributing 8 identical balls into 3 boxes so that no box is empty and each box being large enough to accommodate all balls ?

A. 24
B. 28
C. 36
D. 21
E. 18

Sol. (8-1)C(3-1) = 7C2 = 7*6/2*1 = 21.

2. A group consists of 4 couples  in which each of the 4 men have one wife each. In how many ways could they arranged in a straight line so that the men and women occupy alternate position ?

A. 576
B. 982
C. 1152
D. 1024
E. None of these

Sol. 4!*4! + 4!*4! = 576+576 = 1152.

3. A five digit number is formed with the digits 0,1,2,3 and 4 without repetition.Find the chance that the number is divisible by 2 ?

A. 1/2
B. 2/3
C. 1/3
D. 1/4
E. None of these

Sol. 5 digit number = 5! = 120
Divisible by 2 then the last digit should be 0, 2, 4
Then no. of cases with 0 as last digit = 4! = 24
     no. of cases with 2 as last digit = 18
     no. of cases with 4 as last digit = 18
Total cases = 24 + 18 + 18 = 60
P = 60/120 = 1/2.

4. From a group of 4 men and 3 women , 2 persons are selected at random. Find the probability that at least one woman is selected ?

A. 2/3
B. 3/5
C. 5/7
D. 1/3
E. 1/2

Sol. Total cases = 7c2 = 7!/5!*2! = 21
     Cases when no women is selectd = 4c2 = 4!/2!*2! = 6
     Cases when atleast 1 woman is selected = (21-6) = 15
     Probability = 15/21 = 5/7.

5. 16 persons are participated in a party. In how many different ways can they host the seat in a circular table, if the 2 particular persons are to be seated on either side of the host ?

A. 16! * 2
B. 14! * 2
C. 18! * 2
D. 14!
E. None of these

Sol. (16 – 2)! * 2 = 14! * 2.

6. A and B start a business with investments of Rs. 10000 and Rs. 9000 respectively. After 4 months, A takes out 1/2 of his capital. After 2 more months, B takes out 1/3 of his capital while C joins them with a capital of Rs. 14000. At the end of a year, they earn a profit of Rs. 10160. Find the share of each member in the profit?

A. Rs A – Rs. 3300, B – Rs. 3500, C – Rs. 3360
B. Rs A – Rs. 3200, B – Rs. 3600, C – Rs. 3360
C. Rs A – Rs. 3200, B – Rs. 3700, C – Rs. 3260
D. Rs A – Rs. 3200, B – Rs. 3500, C – Rs. 3460
E. None of these 

Sol. A : B : C = (10,000 x 4 + 5000 x 8) : (9000 x 6 + 6000 x 6) : (14000 x 6)
= 80000 : 90000 : 84000 = 40 : 45 : 42
A’s share = Rs. 10160 x 40/127 = Rs. 3200;
B’s share = Rs. 10160 x 45/127 = Rs. 3600;
C’s share = Rs. 10160 x 42/127 = Rs. 3360.

7. 1, 4, 19, 54, 117, _

A. 194
B. 256
C. 216
D. 200
E. None of these

Sol. 1 + 1*3 = 4
     4 + 3*5 = 19
    19 + 5*7 = 54
    54 + 7*9 = 117
   117 + 911 = 216

8. Prachi started a business investing Rs. 50,000 in 2015, In 2016, she invested an additional amount of Rs. 20,000 and Arsh joined him with an amount of Rs. 70,000. In 2017, Prachi invested another additional amount of Rs. 20,000 and Heena joined them with an amount of Rs. 70,000. What will be Arsh’s share in the profit of Rs. 300,000 earned at the end of 3 years from the start of the business in 2015 ?

A)Rs Rs. 250,000.
B)Rs Rs. 120,000.
C)Rs Rs. 100,000.
D)Rs Rs. 150,000.
E)None of these

Sol. Prachi : Arsh : Heena
= (50000 x 12 + 70000 x 12 + 90000 x 12) : (70000 x 24) : (70000 x 12)
= 2520000 : 1680000 : 840000 = 3 : 2 : 1
Arsh’s share = Rs.300,000 x 2/6 = Rs. 100,000.

9. The ratio of the monthly salaries of A and B is in the ratio 15 : 16 and that of B and C is in the ratio 17 : 18. Find the monthly income of C if the total of their monthly salary is Rs 1,87,450. 

A. Rs 66,240
B. Rs 72,100
C. Rs 62,200
D. Rs 65,800
E. Rs 60,300

Sol. A/B = 15/16 and B/C = 17*18
So A : B : C = 15*17 : 16*17 : 16*18
= 255 : 272 : 288
So C’s salary = [288/(255+272+288)] * 1,87,450

10. X takes 6 days less than Y to finish the work individually. If X and Y working together complete the work in 4 days, then how many days are required by Y to complete the work alone ?

A. 7 days
B. 10days
C. 5 days
D. 12days
E.None of these

Sol. Lets assume Y takes = a days
Then X will take = (a-6) days

Now given, 1/a + 1/(a-6) = 1/4
Solving, we get,
a = 12 days.

Read More

IBPS RRB 2016-17 - Quant Quiz-13

1. (989/34) ÷ (65/869) * (515/207) = ?

A. 845
B. 870
C. 945
D. 745
E. 890

2. 67% of 801 – 231.17 = ? – 23% of 789

A. 400
B. 490
C. 550
D. 600
E. 750

3. (32.13)2 + (23.96)2 – (17.11)2= ?


A. 1410
B. 1310
C. 1550
D. 1650
E. 1810

4. 5907 – 1296 ÷ 144 = ? * 8


A. 700.25
B. 658.25
C. 628.25
D. 737.25
E. 630.5

5. √7378 * √1330 ÷ √660 = ?


A. 150
B. 160
C. 120
D. 170
E. 140

6. 22240 ÷ √? = 34 * 12


A. 3065
B. 3085
C. 3025
D. 3075
E. None of these

7. 8451 + 793 + 620 – ? = 6065 + 713


A. 3486
B. 3586
C. 3286
D. 3186
E. None of these

8. (12.25)2 – √625 = ?


A. 145.1625
B. 125.0625
C. 155.1625
D. 165.0625
E. None of these

9. 156 + 16 * 1.5 – 21 = ?


A. 126
B. 149
C. 141
D. 159
E. None of these

10. (√7921 – √2070.25) * (1/4) = ?


A. 15
B. 16
C. 17
D. 19
E. 11




                                     ANSWERS

1) C. 945
2) B. 490
3) B. 1310
4) D. 737.25
5) C. 120
6) C. 3025
7) E. None of these
8) B. 125.0625
9) D. 159
10) E. 11

Read More

IBPS RRB 2016-17 - Quant Quiz-12 (Based on previous year papers)

1. The average age of some males and 15 females is 18 years. The sum of the ages of 15 females is 240 yrs and average age of males is 20 yrs. Find the number of males ?

A. 10
B. 12
C. 15
D. 18
E. 20

Sol. Lets assume no. of males to be 'x'.

Thus, sum of ages of males/x = 20
      sum of ages of males = 20x

Therefore, (240 + 20x)/(15 + x) = 18
Solving, we get,
x = 15yrs

2. If the sum of the smaller number x and two times the other number is equal to sum of two times the smaller number and 16. The difference between the numbers is 6. Find the smaller number ?

A. 4
B. 5
C. 6
D. 2
E. 3

Sol. Lets assume the larger number be 'y'
Thus, x + 2y = 2x + 16........1st equation
Also, y - x = 6...............2nd eqution

Solving these two, we get,
y = 10
x = 4

3. 16 men and 12 women can complete a work in 8 days, if 20 men can complete the same work in 16 days, in how many days 16 women can complete the same piece of work ?

A. 8
B. 12
C. 10
D. 15
E. 9

Sol. Lets assume it take 'w' days for 1 women to complete the job.
Then, 16/(16*20) + 12/w = 1/8
      w = 160 days
Now for 16 women it will be, = 160/16 = 10 days.

4. If A's salary is 10,000 less than B's salary and B's salary is 15,000 less than C's salary and sum of their salaries is 65,000. Find the salary of A.

A. 15,000
B. 10,000
C. 12,000
D. 20,000
E. 5,000

Sol. Lets assume the salary of C be 'x'
Then, salary of B = x - 15,000
      salary of A = x - 25,000
Now sum of these salaries is,
x + x - 15,000 + x - 25,000 = 65,000
Solving,
x = 35,000
A's salary = 35,000 - 25,000 = 10,000.

5. The sum of the ages of P and Q is 25 yrs more than the age of R.The present age of Q is 5 yrs more than the age of R. Find the present age of P ?

A. 15
B. 20
C. 12
D. 10
E. 16

Sol. Given, P + Q = 25 + R
Also, given, Q = 5 + R
Putting the value of Q in the 1st eqn, we get,
P + 5 + R = 25 + R
P = 20 yrs.

6. The perimeter of the rectangular field is 240 m. The ratio of length and breadth is 8:7. Find the area of the rectangle ?

A. 3854
B. 3584
C. 3485
D. 3845
E. None of these

Sol. Perimeter of the rectangle = 2*(l + b) = 240
Also given, l/b = 8/7, using this in 1st eqn,
8b/7 + b = 120
b = 56
l = 64
Area of the rectangle = 56*64 = 3584 sq.m.

7. One fourth of two-fifth of 30% of a number x is equal to 15. Find 20% of the same number ?

A. 100
B. 120
C. 105
D. 80
E. None of these

Sol. (1/4)*(2/5)*(30/100)*x = 15
     x = 100.

8. The difference between compound interest and simple interest on a sum for 2 yrs at 20% per annum, when the interest is compounded annually is Rs 240. If the interest were compounded half yearly, the difference in two interests over same period would be ?

A. Rs 384.6
B. Rs 344.2
C. Rs 324.8
D. Rs 316.5
E. None of these

Sol. Difference for 2 yrs = P*20*20/100*100 = 240
Solving, we get,

P = 6000

So SI for 2 yrs = 6000*20*2/100 = 2400
Amount at compounded half yearly = 6000 [1 + 10/100]^4

Solving,we get,

Amount at compounded half yearly = 8784.6
So CI at compounded half yearly = 8784.6 – 6000 = 2784.6
So difference = 2784.6 – 2400 = Rs 384.6


Directions(9-10): What should come in place of the question mark(?) in the following number series:

9. 7, 16, 34, 61, 97, ?

A. 142
B. 154
C. 148
D. 164
E. None of these

Sol. 7 + 9*1, 16 + 9*2, 34 + 9*3, 61 + 9*4, ..

10. 8, 2, 2, 4.5, 18, ?

A. 112.5
B. 122.5
C. 134.2
D. 142.5
E. None of these

Sol. *0.5^2, *1^2, *1.5^2, *2^2, *2.5^2,  

Read More