Crack IBPS PO : Algebra New Pattern Questions : Day 43

Directions: 1-5) Read the following  information and answer the questions accordingly.

Q.1) Quantity I: By selling 20 mangoes for Rs.520 a total profit of Rs. 40 is obtained. What is the cost price of one mango?

Quantity II: Total market price of 15 mangoes Rs.500 which was Rs.50 more than the total cost. What is the cost price of one mango?

a)Quantity  I<Quantity II

b) Quantity I>Quantity II

c) Quantity I≤ Quantity II

d) Quantity I≥ Quantity II

e) Quantity I= Quantity II

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By using statement I,

Cost price = 520-40 = 480

Cost price per mango = 480÷20 = Rs.24

By using statement II,

Total cost price = 500-50=450

Cost price of per mango = 450÷15=Rs.30

So, Quantity I< Quantity II

Q.2)

Quantity I : A person marked the price of a refrigerator 60% above the cost price. By giving a discount of 25% the seller sold the refrigerator at the price of Rs.15000 . what is the cost price of the refrigerator?

Quantity II: The cost price of a refrigerator is Rs.7500, which is sold at a profit of 60%. What is the selling price of the refrigerator?

a) Quantity I< Quantity II

b) Quantity I> Quantity II

c) Quantity I≤ Quantity II

d) Quantity I≥ Quantity II

e) Quantity I= Quantity II

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By using statement I

Let the marked price be Rs.x

Then,(x×0.75)=15000

X =Rs.20000

Let cost price be y

1.60×y=20000

Y=Rs.12500

By using statement II

Profit%=profit/(cost price)×100

60%×7500= profit

Profit =Rs.4500

Hence, selling price is Rs.12000

So, Quantity I> Quantity II

Q.3)

Quantity I: Find the average of all the numbers between 1 to 50 that are prime numbers?

Quantity II: Find the average of first 20 natural numbers?

a) Quantity I< Quantity II

b) Quantity I> Quantity II

c) Quantity I≤ Quantity II

d) Quantity I≥ Quantity II

e) Quantity I= Quantity II

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By using statement I

Number of prime numbers between 1 to 50 =15

“Average = (2+3+5+7+11+13+17+19+23+29+31+37+41+43+47) /15

=328 /15  = 21.867

By using statement II

“Average of first 20 natural number s = 20(20+1)/2 =210

=210/20 = 10.5

Hence Quantity I> Quantity II

Q.4)

Quantity I: Pipe A can fill the tank in 40 hours, when it works at 60% of their efficiency. Pipe B is one third as efficient as pipe A. How long will it take if both the pipes operate simultaneously at the 100% efficiency.

Quantity II: 18 hours.

a) Quantity I< Quantity II

b) Quantity I> Quantity II

c) Quantity I≤ Quantity II

d) Quantity I≥ Quantity II

e) Quantity I= Quantity II

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Pipe A can fill the tank when its efficiency is 100% = (40×60)/100=24 hours.

Pipe B can fill the tank in 72 hours.

When both the pipes operate simultaneously, it takes 18 hrs to fill the tank.

Hence answer e is corret.

Q.5)

What is the rate of interest per annum?

Quantity I : The difference between the compound interest and the simple interest accrued on an amount of Rs.50000 at the end of two years is Rs .320

Quantity II: The simple interest after three years for Rs.50000 is Rs.12000

a) Quantity I< Quantity II

b) Quantity I> Quantity II

c) Quantity I≤ Quantity II

d) Quantity I≥ Quantity II

e) Quantity I= Quantity II

Click Here to View Answer

By using statement I,

“Difference between S.I and D.I =”  (PR2)/1002

320=(50000×R2)/1002

Rate of interest = 8

By using statement II

SI = 12000 T = 3 and P = 50000

12000 =  (50000×3×R)/100

R = 8

Hence answer e is correct.

Q.6)Mohan distributed his assets to his wife , four sons, three daughters and six grand children in such a way that each grand child got one-sixteenth of each son and one-tenth of each daughter. His wife got 60% of the total share of his sons and daughter together. If each daughter receives assets of worth Rs.1.25 lakh, what is the share of his wife?

a)Rs. 705000

b)Rs.815000

c) Rs.750000

d) Rs.835000

e) None of these

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“Share of 1 grand child = 1 /10×1.25lakh=0.125 lakhs

Share of 1 son = 16 × 0.125 lakh = 2 lakhs

Share of 4 sons= 4× 2lakhs= 8 lakhs

Share of 3 daughters = 3×1.25 lakhs = 3.75 lakhs

Total share of sons and daughters = (8+3.75)lakhs=11.75 lakhs

6/10×11.75 lakhs=Rs.705000.

Q.7) There are 3 inlet pipes X, Y and Z connected to a tank. If only one pipe is opened at a time, then it takes 50, 40 and 25 minutes for pipes X, Y and Z respectively to fill the tank. Find the time taken to fill 99% of the tank if it is known that in every 5 minutes for the first 2 minutes pipe Y is opened and then closed  for 3 minutes. The remaining pipes are always kept open.

a) 14 minutes

b) 16 minutes

c) 18 minutes

d) 20 minutes

e) Cannot be determined

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Part of the tank filled per minute by pipes X and Z respectively = 2% and 4%

Pipe Y fills 5% of the tank for every 2 minutes it operates.

In 5 minutes, the tank filled by X and Z = 30% and by pipe Y = 5%

So, in 5 minutes , % of tank filled =30+5=35%

In 10 minutes, the tank is filled 70%

For next 2 minutes part of tank filled = 5+12 = 17%

The remaining 12% is filled in time = 2 minutes

Total time taken = 10+2+2 =14minutes

Q.8) A certain number of people get together to contribute in the construction of a charity hospital. But every month four people step out of this plan. Due to this the task is completed in half more year instead of one year. Then how many people were originally involved in this plan?

a) 153

b) 95

c) 102

d) Cannot be determined

e) None of these.

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Let the total number of people = x

Then,

12x = (x+ (x-4)+(x-8)+(x-12)+(x-16)+(x-20)+… 18 times)

12x= 18x-4(1+2+3+…17)

6x=(4×17×18)/2

x=102

Q.9) Find the percentage by which the volume of the circular cylinder change assuming that the radius and the height of the circular cylinder decreases by 20%?

a) 48.8%

b) 46.7%

c) 57.7%

d) 65%

e) None of these

Click Here to View Answer

Volume = π r2 h

Let the radius and height = 10 cm

So area = π×10×10×10 =1000 π cm2

After decrease

New radius = 10-20×  10 /100  = 8 cm

New height = 10-20×  10 /100  = 8 cm

New volume = π×8×8×8 = 512 π cm2

Decresed volume = 488 π cm2

Percentage decrease = 488π×  100 /1000π  =48.8%

Q.10) A cyclist, cycling on a road, passes a man who was walking at the rate of 4 km/hr in the same direction. The man could see the cycle for 12 min and it was visible to him up to a distance of 1.2 km. What was the speed of the cycle?

A) 10 km/hr

b) 12km/hr

c) 6 km/hr

d) Cannot be determined

e) None of these

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Let the speed of cycle be x km/h.

Speed of man = 4 km/h

Relative speed = (x-4) km/h

Therefore, (x-4)×12/60= 1.2

x – 4 = 6

x = 10 km/h


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