35 IBPS RRB Clerk Mocks @ Rs. 149<\/a><\/p>\n 70 IBPS RRB (PO + Clerk) Mocks @ Rs. 199<\/a><\/p>\n Take a free mock test for IBPS RRB Clerk<\/a><\/p>\n Download IBPS RRB Clerk Previous Papers PDF<\/a><\/p>\n Question 1:\u00a0<\/b>There are four members in a family. The ratio of ages of P and R is 5 : 4. The age of Q is 7 less than double the age of P. the age of S is 2 more than double the age of R. The average age of P and Q is same as the average age of R and S. Find the age of S.<\/p>\n a)\u00a024 years<\/p>\n b)\u00a012 years<\/p>\n c)\u00a023 years<\/p>\n d)\u00a026 years<\/p>\n e)\u00a0Can\u2019t be determined<\/p>\n Question 2:\u00a0<\/b>A family consists of four members. The ratio of ages of wife and husband is 2 : 3. The age of Grand father is 150% of the sum of ages of mother and son. The son\u2019s age is one less than 33.33% of the age of father. If the sum of ages of all the family members is 134. Then, find the age of son.<\/p>\n a)\u00a014 years<\/p>\n b)\u00a010 years<\/p>\n c)\u00a012 years<\/p>\n d)\u00a016 years<\/p>\n e)\u00a0Can\u2019t be determined<\/p>\n Question 3:\u00a0<\/b>There are three persons P, Q and R with different ages. If 10 is added to half the age of P and half the age of Q, then the resultant will be equal to Q\u2019s age. If 9 is added to $\\dfrac{1}{4}th$ of Q\u2019s age and half of R\u2019s age, then the resultant will be equal to P\u2019s age. If 1 is added to half of P\u2019s age and $\\dfrac{1}{8}th$ of Q\u2019s age, then the resultant will be equal to R\u2019s age. Then find the age of Q.<\/p>\n a)\u00a032 years<\/p>\n b)\u00a018 years<\/p>\n c)\u00a026 years<\/p>\n d)\u00a024 years<\/p>\n e)\u00a0Can\u2019t be determined<\/p>\n Question 4:\u00a0<\/b>There are three persons A, B and C with different ages. If the age of A is added to half the age of B and twice the age of C, then the result is 105. If the age of B is added to twice the age of A and half the age of C, then the result is 84. If the age of C is added to half the age of A and twice the age of C, then the result is 84. Find the age of C.<\/p>\n a)\u00a036 years<\/p>\n b)\u00a024 years<\/p>\n c)\u00a018 years<\/p>\n d)\u00a012 years<\/p>\n e)\u00a0Can\u2019t be determined<\/p>\n Question 5:\u00a0<\/b>The sum of ages of father and his daughter is 45 years. If the product of their ages 8 years ago was 3 times the age of the father at that time. What will be the age of father after 4 years?<\/p>\n a)\u00a027 years<\/p>\n b)\u00a038 years<\/p>\n c)\u00a011 years<\/p>\n d)\u00a034 years<\/p>\n e)\u00a031 years<\/p>\n Question 6:\u00a0<\/b>Find the approximate value of $\\Large\\frac{169.87^{2}-13.94^{2}}{183.87}$ $\\large+$ $\\Large\\frac{3211.96}{44.002}$<\/p>\n a)\u00a0224<\/p>\n b)\u00a0236<\/p>\n c)\u00a0239<\/p>\n d)\u00a0229<\/p>\n e)\u00a0None of these<\/p>\n Question 7:\u00a0<\/b>Find the approximate value of $17.33+142.895-76.795+235.008-4.779+38.102$<\/p>\n a)\u00a0321<\/p>\n b)\u00a0351<\/p>\n c)\u00a0281<\/p>\n d)\u00a0371<\/p>\n e)\u00a0None of these<\/p>\n Question 8:\u00a0<\/b>Find the approximate value of ‘x’ in $\\Large\\frac{383.96}{x} = \\frac{x}{24.01}$<\/p>\n a)\u00a084<\/p>\n b)\u00a096<\/p>\n c)\u00a092<\/p>\n d)\u00a088<\/p>\n e)\u00a0None of these<\/p>\n Question 9:\u00a0<\/b>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 ?<\/p>\n a)\u00a021.45<\/p>\n b)\u00a021<\/p>\n c)\u00a020<\/p>\n d)\u00a020.45<\/p>\n e)\u00a022.45<\/p>\n Question 10:\u00a0<\/b>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 ?<\/p>\n a)\u00a0340<\/p>\n b)\u00a0350<\/p>\n c)\u00a0360<\/p>\n d)\u00a0370<\/p>\n e)\u00a0380<\/p>\n Free Mock Test for IBPS RRB Clerk<\/a><\/p>\n IBPS RRB Clerk Previous Papers<\/a><\/p>\n Question 11:\u00a0<\/b>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 ?<\/p>\n a)\u00a0175480<\/p>\n b)\u00a0175000<\/p>\n c)\u00a0174480<\/p>\n d)\u00a0175420<\/p>\n e)\u00a0175440<\/p>\n Question 12:\u00a0<\/b>There are 4 people in a family whose ages are P,P+12,P+36 and P+60 years and after \u2018y\u2019 years the oldest of the family leaves the family and the family has only 3 people but the average of the family doesn\u2019t change. After another \u2018y\u2019 years the average of the age of 3 people is 47 years then what is the value of P ?<\/p>\n a)\u00a09<\/p>\n b)\u00a010<\/p>\n c)\u00a011<\/p>\n d)\u00a012<\/p>\n e)\u00a013<\/p>\n Instructions<\/b><\/p>\n Select the correct option.<\/p>\n Question 13:\u00a0<\/b>Quantity 1:A number is increased by 85% and then it is decreased by 15% and result is approximately 357 a)\u00a0Quantity 1 $\\leq$Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 > Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Question 14:\u00a0<\/b>Quantity 1:If the cost price of total apples bought is Rs 288 and each apple is sold at Rs 14 each and the profit percent is 50\/3 then number of apples sold is a)\u00a0Quantity 1 $\\leq$Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 > Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Question 15:\u00a0<\/b>Quantity 1:Numerator of a fraction is increased by 20% an the denominator is decreased by 25% and the resultant is 2 then the sum of numerator and denominator of the original fraction. a)\u00a0Quantity 1 $\\leq$Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 > Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Instructions<\/b><\/p>\n Study the following table and answer the questions that follow. Some values are missing. You need to calculate them as per given information.<\/p>\n Question 16:\u00a0<\/b>If the average marks of B and D in Maths is 102.5, then find the average marks of all the students in Maths<\/p>\n a)\u00a093.5<\/p>\n b)\u00a087.65<\/p>\n c)\u00a097<\/p>\n d)\u00a0102<\/p>\n e)\u00a0Can\u2019t be determined<\/p>\n Question 17:\u00a0<\/b>If the marks obtained by D in Physics is same as the marks obtained by D in Chemistry, then find the ratio between the marks obtained by D in Chemistry and the marks obtained by D in English.<\/p>\n a)\u00a021 : 37<\/p>\n b)\u00a025 : 33<\/p>\n c)\u00a027 : 39<\/p>\n d)\u00a023 : 33<\/p>\n e)\u00a0None of these<\/p>\n Question 18:\u00a0<\/b>If the ratio of marks obtained by C in Physics and marks obtained by B in Physics is 10 : 11, then find the percentage of marks obtained by B in Maths if his overall percentage is 82.2%.<\/p>\n a)\u00a083.33%<\/p>\n b)\u00a074.85%<\/p>\n c)\u00a073.33%<\/p>\n d)\u00a086.5%<\/p>\n e)\u00a0Can\u2019t be determined<\/p>\n Question 19:\u00a0<\/b>If C obtained 81.81% in English and 74% overall, then find the difference between the marks obtained by C and A in chemistry.<\/p>\n a)\u00a012<\/p>\n b)\u00a010<\/p>\n c)\u00a08<\/p>\n d)\u00a03<\/p>\n e)\u00a0Can\u2019t be determined<\/p>\n Question 20:\u00a0<\/b>If the marks obtained by E in Chemistry is 25 more than that in English and if his overall percentage is 64%, then find the marks obtained by E in Chemistry.<\/p>\n a)\u00a070<\/p>\n b)\u00a065<\/p>\n c)\u00a095<\/p>\n d)\u00a084<\/p>\n e)\u00a0Can\u2019t be determined<\/p>\n Quantitative Aptitude formulas PDF<\/a><\/p>\n 520 Banking Mocks – Just Rs. 499<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the ages of P and R be 5x and 4x years respectively. \u21d2 $P+Q = R+S$ Therefore, Age of S = 8x+2 = 8*3+2 = 24+2 = 26 years.<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the ages of wife and husband be 2x and 3x years respectively. Age of grandfather = 150% of 2x+x-1 = $\\dfrac{3}{2}\\times(3x-1) = \\dfrac{9x-3}{2}$ years<\/p>\n Given, $2x+3x+x-1+\\dfrac{9x-3}{2} = 134$<\/p>\n \u21d2 $6x+\\dfrac{9x-3}{2} = 135$<\/p>\n => $21x-3 = 270$ Therefore, Age of Son = $x-1 = 13-1 = 12$ years<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the present ages of P, Q and R be P, Q and R years respectively. \u21d2 $P+R+20 = 2Q$ $\\dfrac{Q}{4}+\\dfrac{R}{2}+9 = P$<\/p>\n \u21d2 $Q+2R+36 = 4P$ $\\dfrac{P}{2}+\\dfrac{Q}{8}+1 = R$<\/p>\n \u21d2 $4P+Q+8 = 8R$ Solving (1) and (2), Solving (2) and (3), Solving (4) and (5), 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the ages of A, B and C be A, B and C years respectively. \u21d2 $2A+B+4C = 210$ \u2192 (1)<\/p>\n $2A+B+\\dfrac{C}{2} = 84$<\/p>\n \u21d2 $4A+2B+C = 168$ \u2192 (2)<\/p>\n $\\dfrac{A}{2}+2B+C = 84$<\/p>\n \u21d2 $A+4B+2C = 168$ \u2192 (3)<\/p>\n Solving (1) and (3) 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the present age of father be \u2018f\u2019 and the daughter be \u2018d\u2019 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $\\Large\\frac{169.87^{2}-13.94^{2}}{183.87}$ $\\large+$ $\\Large\\frac{3211.96}{44.002}\\simeq \\frac{170^{2}-14^{2}}{184}$ $\\large+$ $\\Large\\frac{3212}{44}$<\/p>\n $= \\Large\\frac{(170+14)(170-14)}{184}$ $+$ $73$<\/p>\n $= \\Large\\frac{184\\times156}{184}$ $+$ $73$<\/p>\n $= 156+73 = 229$<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $17.33+142.895-76.795+235.008-4.779+38.102 \\simeq 17+143-77+235-5+38$<\/p>\n Adding all positive terms: Adding all negative terms: $\\therefore$ $17+143-77+235-5+38 = 433-82 = 351$<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n <\/p>\n $\\Large\\frac{383.96}{x} = \\frac{x}{24.01}$<\/p>\n $\\Rightarrow x^{2} \\simeq 384\\times24 = 9216$<\/p>\n $\\therefore$ x $= \\sqrt{9216} = 96$<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n 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 29,28,27\u2026…12 10)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Given average for first 85 innings is 52 11)\u00a0Answer\u00a0(E)<\/strong><\/p>\n let the four numbers be a,b,c and d 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Present average=(P+P+12+P+36+P+60)\/4 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n In Q1 we have x*1.85*0.85=357 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n In Q1 we have Total CP=288 15)\u00a0Answer\u00a0(E)<\/strong><\/p>\n In Q1 let the fraction be be (x\/y) 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given, average marks of B and D in Maths is 102.5 Marks obtained by A in Maths = 46.67% of 150 = $\\dfrac{7}{15}\\times150 = 70$<\/p>\n Marks obtained by C in Maths = 66.67% of 150 = $\\dfrac{2}{3}\\times150 = 100$<\/p>\n Marks obtained by E in Maths = 73.33% of 150 = $\\dfrac{11}{15}\\times150 = 110$<\/p>\n Therefore, Required average = $\\dfrac{70+100+110+205}{5} = \\dfrac{485}{5} = 97$<\/p>\n 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Marks obtained by D in Chemistry = Marks obtained by D in Physics = 62.5% of 120 = $\\dfrac{5}{8}\\times120 = 75$ Therefore, Required ratio = 75 : 99 = 25 : 33.<\/p>\n 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Marks obtained by C in Physics = 83.33% of 120 = $\\dfrac{5}{6}\\times120 = 100$<\/p>\n Then, Marks obtained by B in Physics = 110 ( C\u2019s Physics marks : B\u2019s Physics marks = 10 : 11) Marks obtained by B in English = 80% of 110 = 88 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Marks obtained by C in English = 81.81% of 110 = $\\dfrac{9}{11}\\times110 = 90$<\/p>\n Marks obtained by C in Maths = 66.67% of 150 = $\\dfrac{2}{3}\\times150 = 100$<\/p>\n Marks obtained by C in Physics = 83.33% of 120 = $\\dfrac{5}{6}\\times120 = 100$<\/p>\n Total marks obtained by C = 74% of 500 = 370<\/p>\n Then, Marks obtained by C in Chemistry = 370 – (90+100+100) = 80<\/p>\n Marks obtained by A in Chemistry = 58.33% of 120 = $\\dfrac{7}{12}\\times120 = 70$<\/p>\n Therefore, Difference = 80 – 70 = 10.<\/p>\n 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Marks obtained by E in Maths = 73.33% of 150 = $\\dfrac{11}{15}\\times150 = 110$<\/p>\n Marks obtained by E in Physics = 37.5% of 120 = $\\dfrac{3}{8}\\times120 = 45$<\/p>\n Let marks obtained by E in English = x% of 110 = 1.1x
\nQuantity 2:A number is increased by 82% and then it is decreased by 12% and the result is approximately 355.5<\/p>\n
\nQuantity 2:An article is marked up by 40% an a discount of 30% was given then 13 times the loss percent is<\/p>\n
\nQuantity 2:Numerator of a fraction is increased by 50% and denominator is decreased by 60% and the resultant is 3 then the sum of numerator and denominator of original fraction is<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/DI18.JPG\")
<\/figure>\n
\nAge of S = 2(4x)+2 = 8x+2 years
\nAge of Q = 2(5x)-7 = 10x-7 years
\nGiven, $\\dfrac{P+Q}{2} = \\dfrac{R+S}{2}$<\/p>\n
\n\u21d2 $5x+10x-7 = 4x+8x+2$
\n\u21d2 $15x-7 = 12x+2$
\n\u21d2 $3x = 9$
\n\u21d2 $x = 3$<\/p>\n
\nAge of son = $\\dfrac{3x}{3}-1 = x-1$ years.<\/p>\n
\n\u21d2 $21x = 273$
\n\u21d2 $x = 13$<\/p>\n
\n$\\dfrac{P}{2}+\\dfrac{R}{2}+10 = Q$<\/p>\n
\n\u21d2 $P-2Q+R = -20$ \u2192 (1)<\/p>\n
\n\u21d2 $4P-Q-2R = 36$ \u2192 (2)<\/p>\n
\n\u21d2 $4P+Q-8R = -8$ \u2192 (3)<\/p>\n
\n$6P-5Q=-4$ \u2192 (4)<\/p>\n
\n$-12P+5Q=-152$ \u2192 (5)<\/p>\n
\nQ = 32
\nTherefore, The present age of Q = 32 years.<\/p>\n
\n$A+\\dfrac{B}{2}+2C = 105$<\/p>\n
\n7B = 126 \u21d2 B = 18
\nSubstituting B value in (2) and (3)
\n(2) \u21d2 4A+36+C = 168
\n\u21d2 4A+C = 132 \u2192 (4)
\n(3) \u21d2 A+72+2C = 168
\n\u21d2 A+2C = 96 \u2192 (5)
\nSolving (4) and (5), we get
\nA = 24 and C = 36
\nTherefore, The present ages of A, B and C are 24 years, 18 years and 36 years respectively.<\/p>\n
\nAs per the given question,
\nf + d = 45 \u2026.(1) and
\n(f – 8)(d – 8) = 3(f – 8)
\nSubstitute equation (1) in the above equation
\n(f – 8)(37 – f) = 3(f – 8)
\n37 – f = 3 (or) f = 34
\nThe age of the father after 4 years will be 34 + 4 = 38 years.
\nHence, option B is the correct answer.<\/p>\n
\n$17+143+235+38 = 433$<\/p>\n
\n$77+5 = 82$<\/p>\n
\nSo the total number of passengers=30+30+29+….12
\n=30+(30(31)\/2)-(11(12)\/2)
\n=30+465-66
\n=399+30
\n=429
\nTotal number of compartments=20
\nAverage=429\/20
\n=21.45<\/p>\n
\nSum of the runs scored by Sehwag for first 85 innings=85*52
\n=4420
\nSum of the runs scored by him in first 90 innings =90*53
\n=4770
\nRuns scored by him from 86 to 90 innings=4770-4420
\n=350
\nRuns scored by him in first 100 innings=100*55
\n=5500
\nRuns scored by him in last 10 innings=5500-4770
\n=730
\nRequired difference=730-350
\n=380<\/p>\n
\nGiven a+b+c+d=147*4
\na+b+c+d=588
\na=5b\/17
\nb=c-36
\nc=b+36
\nc=(d\/2)+52
\nb+36=(d+104)\/2
\n2b+72=d+104
\nd=2b-32
\n(5b\/17)+b+b+36+2b-32=588
\n(5b\/17)+4b+4=588
\n((5b+68b)\/17) =584
\n73b=584*17
\nb=136
\na=40
\nc=172
\nd=240
\nLCM of 40,136,172,240 is 175440<\/p>\n
\n=(4P+108)\/4
\nAfter \u2018y\u2019 years P+60 age person will leave so now average becomes
\n((P+y+P+12+y+P+36+y)\/3)=(4P+108)\/4
\n(3P+3y+48)\/3 =P+27
\nP+y+16=P+27
\ny=27-16
\ny=11
\nNow (3P+6y+48)\/3 =47
\n3P+6y+48=141
\n3P=141-48-66
\n3P=27
\nP=9<\/p>\n
\nx=357\/(1.85*0.85)
\nx=227
\nIn Q2 we have x*1.82*0.88=355.5
\nx=355.5\/(1.82*0.88)
\nx=220
\nTherefore Q1>Q2<\/p>\n
\nLet the number of apples be x
\nTotal SP=14x
\nTherefore ((14x-288)\/(288))*100=50\/3
\n14x-288=48
\n14x=288+48
\nx=336\/14
\nx=24
\nIn Q2 we have CP=100
\nMP=100*1.4=140
\nSP=MP-discount
\nSP=140-140*(30\/100)
\nSP=98
\nLoss percent=((100-98)\/(100))*100
\n=2%
\n13 times the loss percent=26
\nTherefore Q2 > Q1<\/p>\n
\nThen as given (1.2x)\/(0.75y)=2
\nx\/y =5\/4
\nsum=5+4=9
\nIn Q2 let the fraction be (x\/y)
\nThen as given (1.5x)\/(0.4y)=3
\nx\/y =4\/5
\nsum=4+5=9<\/p>\n
\nThen, Total marks obtained by B and D in Maths = 205.<\/p>\n
\nMarks obtained by D in English = 90% of 110 = 99<\/p>\n
\nMarks obtained by B in Chemistry = 73.33% of 120 = $\\dfrac{11}{15}\\times120 = 88$<\/p>\n
\nOverall percentage of B = 82.2%
\nTotal marks obtained by B = 82.2% of 500 = 411
\nThen, Marks obtained by B in Maths = 411 – (110+88+88) = 125
\nTherefore, Required percentage = $\\dfrac{125}{150}\\times100 = \\dfrac{5}{6}\\times100 = 83.33$%.<\/p>\n
\nThen, Marks obtained by E in Chemistry = 1.1x+25
\nOverall percentage of E = 64%
\nTotal marks obtained by E = 64% of 500 = 320.
\n110+45+1.1x+1.1x+25 = 320
\n=> 2.2x = 140
\n=> x = 63.63
\nThen, Marks obtained by E in Chemistry = 1.1x+25 = $1.1\\times63.63 + 25 = 70+25 = 95$<\/p>\n