# Expected Maths Questions For IBPS RRB PO

Download Top-20 IBPS RRB PO Expected Maths Questions PDF. Expected Maths questions based on asked questions in previous year exam papers very important for the IBPS RRB PO (Officer Scale-I, II & III) exam.

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**Question 1:Â **Find the approximate value of $\Large\frac{169.87^{2}-13.94^{2}}{183.87}$ $\large+$ $\Large\frac{3211.96}{44.002}$

a)Â 224

b)Â 236

c)Â 239

d)Â 229

e)Â None of these

**Question 2:Â **Find the approximate value of $17.33+142.895-76.795+235.008-4.779+38.102$

a)Â 321

b)Â 351

c)Â 281

d)Â 371

e)Â None of these

**Question 3:Â **Find the approximate value of ‘x’ in $\Large\frac{383.96}{x} = \frac{x}{24.01}$

a)Â 84

b)Â 96

c)Â 92

d)Â 88

e)Â None of these

**Question 4:Â **Find the approximate value of $\sqrt{6560}\div2.98^{2}+\sqrt[3]{2189}\div\sqrt{2700}\times\sqrt{5180}\div\sqrt[3]{25}$

a)Â 11

b)Â 15

c)Â 20

d)Â 5

e)Â None of these

**Question 5:Â **Find the approximate value of $153.001\times146.95-66.009\times73.952+25.001\times23.856$

a)Â 17952

b)Â 18207

c)Â 18456

d)Â 17605

e)Â None of these

**Question 6:Â **Â Find the approximate value of $\small1798.98\div8.01\times51.965\div779.97-390.001\div25.87$

a)Â 5

b)Â 12

c)Â 0

d)Â 20

e)Â None of these

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**Question 7:Â **There are 20 compartments in a train. If each compartment can contain upto maximum of 30 people and each compartment should have different number of people except 2 compartments and no compartment should have less than 12 people in it. If the maximum number of passengers are to be filled in the train then what is the average number of people in each compartment ?

a)Â 21.45

b)Â 21

c)Â 20

d)Â 20.45

e)Â 22.45

**Question 8:Â **Average runs scored by Sehwag in his first 85 innings was 52 and in each of the next 5 innings till 100 innings his average gradually increased as 53 and then to 54 and then to 55 at the end of 100 innings. What is the difference between the runs scored by him in 86 to 90 innings and the last 10 innings ?

a)Â 340

b)Â 350

c)Â 360

d)Â 370

e)Â 380

**Question 9:Â **Average of four numbers is 147. First number is (5/17) times the second number and the second number is 36 less than the third number and the third number is 52 more than the half of the fourth number. What is the LCM of four numbers ?

a)Â 175480

b)Â 175000

c)Â 174480

d)Â 175420

e)Â 175440

**Question 10:Â **There are 4 people in a family whose ages are P,P+12,P+36 and P+60 years and after â€˜yâ€™ years the oldest of the family leaves the family and the family has only 3 people but the average of the family doesnâ€™t change. After another â€˜yâ€™ years the average of the age of 3 people is 47 years then what is the value of P ?

a)Â 9

b)Â 10

c)Â 11

d)Â 12

e)Â 13

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**Question 11:Â **A circle is inscribed inside a rectangle of length 12 cm and breadth Y cm. The circle touches the inner part of the breadth of the rectangle. If the difference between their areas is $\dfrac{320}{7}$ sq.cm, then find Y.

a)Â 6 cm

b)Â 8 cm

c)Â 4.75 cm

d)Â 7.84 cm

e)Â 9 cm

**Question 12:Â **The area of a rectangle is 68 sq.cm. If the length of the rectangle is reduced by 3 cm and the breadth of the rectangle is increased by 5 cm, then its area reduces to 12 less than half of its original area. If its length is increased by 2 cm and its breadth is reduced by 4 cm, then its area will be 10 more than its original area. Find its length.

a)Â 7.65 cm

b)Â 3.42 cm

c)Â 4 cm

d)Â 6 cm

e)Â None of these

**Question 13:Â **If two sides of a triangle are 6 cm and 4 cm then how many integral values are possible for the third side ?

a)Â 6

b)Â 7

c)Â 8

d)Â 9

e)Â 10

**Question 14:Â **A person standing on the top of a building of height 20m observes the angle of elevation of the top of other building as 60 degrees and angle of depression of the bottom the same building as 30 degrees. What is the difference between the heights of two building ?

a)Â 10m

b)Â 40m

c)Â 50m

d)Â 30m

e)Â 60m

**Question 15:Â **A cuboid has its length, breadth and height in the ratio of 4:5:6 If the length is decreased b 25% ,breadth is increased by 20% and height is increased by 33.33 % then what is the percentage change in total surface area of the cuboid ?

a)Â 21.1%

b)Â 23.8%

c)Â 22.3%

d)Â 21.6%

e)Â 22.6%

**Question 16:Â **Ram invested in stocks in three investments in the ratio 6:8:15 and he got 20% profit on 1st investment and 37.5% loss in second investment and 10% profit in the third investment. Find his overall profit or loss percentage.

a)Â 20.36%

b)Â 19.25%

c)Â 21.78%

d)Â 19.65%

e)Â 21.37%

**Question 17:Â **A man has 2 sons and the ages of them are 12 years and 15 years.He deposits total of Rs 3,60,000 in the bank for a time period that is until any one of them reaches 18 years and the money is deposited in the ratio of their present ages.What is the difference between the total amount they receive after the stipulated time if rate of interest is 5% simple interest ?

a)Â 42000

b)Â 46000

c)Â 40000

d)Â 32000

e)Â 36000

**Question 18:Â **A person purchases 200 apples at Rs 300 per dozen.He sold 40% of the apples at Rs 30 per each apple and 25% of the remaining apples were spoiled and in the remaining apples half were sold at Rs 40 per each and the other half were sold at 35 per each. Find the profit/loss percent

a)Â 15.5%

b)Â 14%

c)Â 15.7%

d)Â 14.8%

e)Â 15%

**Question 19:Â **A trader got 10%,20% and 30% profit by selling three article A,B and C respectively.If the sum of cost price by A and B is twice that of C and the difference between the cost price of A and B(A>B) is 40% of the cost price of A. What is the overall profit of the trader ?

a)Â 60.2%

b)Â 60.8%

c)Â 60.4%

d)Â 60.6%

e)Â 70%

**Question 20:Â **To meet the demand during summer, a milkman mixed water with pure milk in 2:5. Then he sold the same mixture at 20% more than normal price. By what percent his revenue will increase during summers? (Water is available without any cost)

a)Â 60 percent

b)Â 50 percent

c)Â 40 percent

d)Â 68 percent

e)Â 63 percent

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**Answers & Solutions:**

**1)Â AnswerÂ (D)**

$\Large\frac{169.87^{2}-13.94^{2}}{183.87}$ $\large+$ $\Large\frac{3211.96}{44.002}\simeq \frac{170^{2}-14^{2}}{184}$ $\large+$ $\Large\frac{3212}{44}$

$= \Large\frac{(170+14)(170-14)}{184}$ $+$ $73$

$= \Large\frac{184\times156}{184}$ $+$ $73$

$= 156+73 = 229$

**2)Â AnswerÂ (B)**

$17.33+142.895-76.795+235.008-4.779+38.102 \simeq 17+143-77+235-5+38$

Adding all positive terms:

$17+143+235+38 = 433$

Adding all negative terms:

$77+5 = 82$

$\therefore$ $17+143-77+235-5+38 = 433-82 = 351$

**3)Â AnswerÂ (B)**

$\Large\frac{383.96}{x} = \frac{x}{24.01}$

$\Rightarrow x^{2} \simeq 384\times24 = 9216$

$\therefore$ x $= \sqrt{9216} = 96$

**4)Â AnswerÂ (B)**

$\sqrt{6560}\div2.98^{2}+\sqrt[3]{2189}\div\sqrt{2700}\times\sqrt{5180}\div\sqrt[3]{25} \simeq \sqrt{6561}\div3^{2}+\sqrt[3]{2197}\div\sqrt{2704}\times\sqrt{5184}\div\sqrt[3]{27}$

$= \Large\frac{\sqrt{6561}}{3^{2}}+\frac{\sqrt[3]{2197}\times\sqrt{5184}}{\sqrt{2704}\times\sqrt[3]{27}}$

$= \Large\frac{81}{9}+\frac{13\times72}{52\times3}$ $= 9+6 = 15$

**5)Â AnswerÂ (B)**

$153.001\times146.95-66.009\times73.952+25.001\times23.856 \simeq 153\times147-66\times74+25\times24$

$153\times147$ can be written as $(150+3)(150-3)$

$=$ $150^{2}-3^{2}$ $(\because (a+b)(a-b) = a^{2}-b^{2})$

$=$ $22500-9$ $=$ $22491$

$66\times74$ can be written as $(70-4)(70+4)$ $=$ $70^{2}-4^{2}$ $=$ $4900-16 = 4884$

$25\times24$ can be written as $24\times\Large\frac{100}{4} =$ $600$

$\therefore$ $153\times147-66\times74+25\times24 = 22491-4884+600 = 18207$

**6)Â AnswerÂ (C)**

$1798.98\div8.01\times51.965\div779.97-390.001\div25.87 \simeq 1800\div8\times52\div780-390\div26$

$= \Large\frac{1800\times52}{8\times780}-\frac{390}{26}$

$= 15-15 = 0$

**7)Â AnswerÂ (A)**

It is given that in each compartment maximum of 30 people can be accommodated and only two compartments can have the same number of people and maximum number of people have to be accommodated in the train and so 30 people should be in two compartments and in the other compartments we should have 29,28,27â€¦…12

So the total number of passengers=30+30+29+….12

=30+(30(31)/2)-(11(12)/2)

=30+465-66

=399+30

=429

Total number of compartments=20

Average=429/20

=21.45

**8)Â AnswerÂ (E)**

Given average for first 85 innings is 52

Sum of the runs scored by Sehwag for first 85 innings=85*52

=4420

Sum of the runs scored by him in first 90 innings =90*53

=4770

Runs scored by him from 86 to 90 innings=4770-4420

=350

Runs scored by him in first 100 innings=100*55

=5500

Runs scored by him in last 10 innings=5500-4770

=730

Required difference=730-350

=380

**9)Â AnswerÂ (E)**

let the four numbers be a,b,c and d

Given a+b+c+d=147*4

a+b+c+d=588

a=5b/17

b=c-36

c=b+36

c=(d/2)+52

b+36=(d+104)/2

2b+72=d+104

d=2b-32

(5b/17)+b+b+36+2b-32=588

(5b/17)+4b+4=588

((5b+68b)/17) =584

73b=584*17

b=136

a=40

c=172

d=240

LCM of 40,136,172,240 is 175440

**10)Â AnswerÂ (A)**

Present average=(P+P+12+P+36+P+60)/4

=(4P+108)/4

After â€˜yâ€™ years P+60 age person will leave so now average becomes

((P+y+P+12+y+P+36+y)/3)=(4P+108)/4

(3P+3y+48)/3 =P+27

P+y+16=P+27

y=27-16

y=11

Now (3P+6y+48)/3 =47

3P+6y+48=141

3P=141-48-66

3P=27

P=9

**11)Â AnswerÂ (B)**

Given, Length of the rectangle = 12 cm

Breadth of the rectangle = Y cm

Area of the rectangle = 12Y sq.cm

The circle touches the inner part of the breadth of the rectangle.

Then, Diameter of the circle = Breadth of the rectangle

Radius of the circle = $\dfrac{Y}{2}$ cm

Area of the circle = $\dfrac{22}{7}\times\dfrac{Y}{2}\times\dfrac{Y}{2} = \dfrac{11Y^2}{14} sq.cm$

Given, $12Y – \dfrac{11Y^2}{14} = \dfrac{320}{7}$

=> $168Y – 11Y^2 = 640$

=> $11Y^2-168Y+640 = 0$

Solving above equation, we get,

Y = 8 or Y = 7.27.

Therefore, The breadth of the rectangle can be either 8 cm or 7.27 cm.

Hence, From the options given, Breadth of the rectangle = 8 cm

**12)Â AnswerÂ (C)**

Let the original length of the rectangle be l cm and the original breadth of the rectangle be b cm.

Then its area = lb = 68 sq,cm

If the length is reduced by 3 cm and breadth is increased by 5 cm,

(l-3)(b+5) = 68/2-13 = 22

lb+5l-3b-15 = 22

68+5l-3b-15 = 22

5l-3b+53 = 22

5l-3b+31 = 0 — (1)

If the length is increased by 2 and breadth is reduced by 4,

(l+2)(b-4) = 68+10 = 78 sq.cm

lb-4l+2b-8 = 78

-4l+2b = 18

4l-2b+18 = 0 — (2)

Solving (1) and (2)

2l – 8 = 0

l = 4

Therefore, Original length of the rectangle = 4 cm.

**13)Â AnswerÂ (B)**

For a triangle to form the sum of two sides should be greater than the third side and the difference of two sides should be less than the third side

Therefore (6+4)>s>(6-4)

10>s>2

So all the values between 2 and 10 are possible as third side

i.e 3,4,5,6,7,8 and 9

Therefore 7 values are possible

**14)Â AnswerÂ (E)**

Let the height of other building be â€˜hâ€™

Distance between the buildings be â€˜dâ€™

From the figure we have $\tan60$=d/20

d=$\sqrt{3}20$

From the figure we have in the upper triangle $\tan60$=x/d

$\sqrt{3}$=x/$\sqrt{3}20$

x=60 cm

Therefore h=60+20

h=80 cm

Difference=80-20

=60 cm

**15)Â AnswerÂ (D)**

Let the length,breadth and height of the cuboid be 4x,5x and 6x respectively

Total surface area of the cube=2lb+2bh+2lh

=$2*(20*x^{2})+2*(24*x^{2})+2*(30*x^{2})$

=$(40+48+60)x^{2}$

=$148x^{2}$

New length of cuboid=4x*3/4

=3x

New breadth of cuboid=5x*(120/100)

=6x

New height of cuboid=6x*(4/3)

=8x

Total surface area of cuboid with new dimensions=$2*(18*x^{2})+2*(48*x^{2})+2*(24*x^{2})$

=$(36+96+48)x^{2}$

=$180x^{2}$

Percentage change in the total surface area=((180-148)/148)*100

=800/37

=21.6%

**16)Â AnswerÂ (D)**

Let the three investments be 6x,8x and 15x.

Given on 1st investment profit=20%

His selling price=6x*1.2

=7.2x

Given on 2st investment loss=37.5%

His selling price=8x*11/8

=11x

Given on 3rd investment profit=10%

His selling price=15x*1.1

=16.5x

Total CP=29x

Total SP=7.2x+11x+16.5x

=34.7x

Profit percent=((34.7x-29x)/29x)*100

=19.65%

**17)Â AnswerÂ (B)**

From the given information the time period is 18-15=3 years

So the money is distributed in the ratio 4:5 i.e On 1st son it is (4/9)*360000=160000

On the second son it is (5/9)*360000=200000

Interest for 1st son=160000*3*5/100

=24000

Interest for 2n son=200000*3*5/100

=30000

Amount received by 1st son=160000+24000

=184000

Amount received by 2nd son=200000+30000

=230000

Difference=230000-184000

=46000

**18)Â AnswerÂ (A)**

Given CP per dozen=300

For each apple cost=300/12

=Rs 25

Total cost price of apples=200*25

=5000

He sold 40% of apples at Rs 30 per each

So (40/100)*200 =80 apples

80*30=Rs 2400

25% of remaining apples are spoiled i.e (1/4)*120=30 apples

So total leftover apples=120-30

=90

So 45 apples are sold at 40 per each and the other 45 apples at 35 per each

Therefore total SP=45*40+45*35

=1800+1575

=Rs 3375

Total SP=3375+2400

=5775

Profit percent=((5775-5000)/(5000))*100

=775/50

=15.5 %

**19)Â AnswerÂ (B)**

Let the cost price of A,B and C be a,b and c

SP of A=1.1a

SP of B=1.2b

SP of C=1.3c

Total SP=1.1a+1.2b+1.3c

Total CP=a+b+c

Aso given a+b=2c

a-b=0.4a

b=0.6a

a+b=2c

1.6a=2c

c=0.8a

Total SP=1.1a+1.2(0.6a)+1.3(0.8a)

=1.1a+0.72a+1.04a

=3.86a

Total CP=a+0.6a+0.8a

=2.4 a

Total profit percent=((3.86a-2.4a)/(2.4a))*100

=60.8%

**20)Â AnswerÂ (D)**

Let us assume that on a normal day he sells milk at x rupee per litre and he sells y litre of milk

So revenue generated on normal day = x*y = xy

He mixed water with milk in ratio of 2:5 so net quantity = $\frac{2+5}{5} * y$ = 1.4y

New selling price during summer = 1.2x rupees per litre

So revenue generated in summers = 1.4y * 1.2x = 1.68xy

Hence we can say that revenue will go up by 68 percent.