IBPS PO Practice Mock Test for - 2 Free Online Test 2026

Apply BODMAS Rule

('of' is equivalent to 'X')

Simplifying both sides

X=13/22

=

Factorizing 74088 = 7 × 7 × 7 × 3 × 3 × 3 × 2 × 2 × 2

7 x 3 x 2 = 42

export is more than import => export to import ratio is greater than 1.From the graph it is clear that for company A, in the years 1995,1996 and 1997 export is more than import.

Only ratio is given. For getting the difference amount of export or import is needed. So ,difference cannot be determined.

Let the amount of A imports of Company A in 1998 be Rs.x crores  Then, (237/x )= 0.75 => x = 316.

In 1997 for Company A we have :E/I= 1.75 => E=1.75I(read E=Export and I=Import) Now required import I’=I+I *40/100 =1.4I New ratio E/I’ = 1.75I / 1.4I =1.25

In 1995 for company !,E!/I!=1.75………..(i)

For company B, EB/IB=0.75………….(ii) Given EA=2EB.............................................(iii) Substitute IA=Rs.180 crores in  (i) we get EA= Rs.315 crores.

Substituting value of EA  in (iii) we get

EB=Rs. 315/2 crores

IB= EB/ 0.75 = Rs.210 crores.

Gold in C=( 7/9 + 7/18) units=7/6 units.
Copper in C= ( 2/9 + 11/18) units=5/6 units.
Gold: Copper=7/6:5/6=7:5

Possible out comes(E) or Favourable outcomes are

E1.2 red ball             or

E2.2 green ball            or

E3.1 red and 1 green

Number of selecting 2 balls from two red balls n(E)=2C2=1

Number of ways of selecting 2 balls from 3 green balls n(E2)= 3C2=3

Number of ways of selecting 1 red ball and 1 green ball n(E3)= 2C1*3C1=2*3=6

Sample space S=Number of ways of selecting 2 balls from 7(2+3+2) balls n(S)=7C2=21

n(E)=n(E1)+ n(E2)+ n(E3)=1+3+6=10

Required probability P(E)=n(E)/n(S)=10/21

Total number of possible ways selecting 2 card from 52 cards is 

n(S)=52C2=(52*51)/(2*1)=1326

Let A be event of getting both black card.

B be event of getting both queens.

Event of getting two black queens A∩B

There are 26 black cards in a pack and 4 queens. Two queens are black

n(A)=Number of ways of selecting  2 blacks from 26 cards=26C2=325

n(B)=Number of ways of selecting 2 queens from 4 cards =4C2=6

n(A∩B)=Chances of getting 2 black queens=2C2=1

P(A)=325/1326
P(B)=6/1326
P(A∩B)=1/1326
P(A∪B)=P(A)+P(B)-P(A∩B)

=(325/1326)+(6/1326)-(1/1326) = 330/1326 =55/221

There are 8 different letters in the given word, but it is given that vowels should come together.

Treat vowels as a single entity ‘AUE’ So total number of letters is 5 different letters+ vowel entity=6

Total arrangement possible using 6 letters=6P6=6!=720

Vowels can be arranged in 3P3 ways. 3P3=6

Total number of arrangement=720*6=4320

Total sales of branches B1, B3 and B5 for both the years (in thousand numbers) = (80 + 105) + (95 + 110) + (75+ 95) = 560.

 Required Percentage =  [{(70+80)/(95+110)} *100]% = 73.17%.

Average sales of all the ,six branches (in thousand numbers) for the year 2000 = (1/6)*[80+ 75 +95 + 85+ 75 + 70] = 80.

Required [(75+65)/(85+95)]=7/9

Average sales (in thousand numbers) of branches B1, B3 and B6 in 2000 =(1/3)*(8O+95+7O)=245/3

Average sales (in thousand numbers) of branches B1, B2 and B3 in 2001 (1/3)*(105+65+110) =280/3

Required Percentage =[(245/3)/(280/3)]*100% = 87.5%

Let distance from R to P be x km.

Then x/32 + x/72 = 13 Solving x we get x=288

Work done by first pipe in 1 hr= 1/14

Work done by second pipe in 1 hr= 1/16

Work done by both pipes in 1 hr= 1/14 + 1/16 =15/112

Time taken by the pipes to fill the cistern(without leakage)= 112/15 = 7 hr 28 mts.

Actual time taken by the pipes to fill the cistern(including leakage)= 7 hr 28 mts+ 32 mts = 8 hrs.

 Work done by 2 pipes and leakage in 1 hr = 1/8

Work done by leak in 1 hr = 15/112 – 1/8 = 1/112

Time taken by the leak to empty cistern = 112 hrs

Profit of Rs.600 at 5% per annum will be on investment= 600*(100/5)= 12000.

Therefore total investment of B for both years= Rs.12000.

Let investment of B in first year is x ,then x + 2x =12000 => x=4000

Investment of B for first year = 4000 and that of second year=8000.

Investment of A for first year= 4 * investment of b for first year= 4*4000=16000.

Investment of B for second year= 16000-2000=14000.

Total investment of A for two years= 16000+14000=30000.

So, profit of A= 30000*(5/100)=1500.

C.P of wire= 500(75/100)= Rs.375

C.P of 60 % wire = 375*(60/100)=Rs.225

Profit made on 60% wire= (8/100) *225=Rs.18

S.P of 60% of wire= Rs.225+Rs.18= Rs.243

Required gain on whole business=12%

Therefore ,S.P of whole wire = (12/100)*375=Rs.420

So S.P of  remaining 40% wire = Rs.420-Rs.243=Rs.177

C.P of 40% = (40/100)*375 = Rs.150

So ,required gain percentage on 40% wire = [(177-150)/150]*100 =18%

Let x be the original rate

Sugar that can be purchased for Rs.140 at rate Rs.x = 140/x  gm

Reduced rate =90x/100

Sugar that can be purchased for Rs.140 at rate Rs.(90x/100) = 140/(90x/100) gm 140/(90x/100)- 140/x   = 500

x= Rs.0.031111 per gm= Rs.(1000*0.031111) per kg= Rs. 31.11

Let the sum be P

S.I for 2 years= PNR/100 =( P*2*10)/100= P/5

C.I for 2 years = P[(1+R/100)2 -1]=P[ (1+10/100)2-1]=21P/100

C.I-S.I=(21P/100 )-(P/5)=60

Solving for P we get P=Rs.6000.

Total age of 34 boys =34*14=476

Total age of 34 boys +teacher = 35*15=525

Age of teacher = 525-476=49 years.

Let the age of son be x, and that of father be y. y-4=8(x-4)……..(1)

y=4x………………(2)

Solving eqn (1) and (2)

x= 7

y=28

C.P of Vinu= (9/10)S.P S.P of vinu =

% profit= {[(9/50) S.P ]/[ (9/10)S.P] } * 100 = 20%